leecode更新
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dingjiawen
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48fa58f8f2
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bb316cd55c
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package com.markilue.leecode.string;
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import org.junit.Test;
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import java.util.Arrays;
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/**
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* @BelongsProject: Leecode
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* @BelongsPackage: com.markilue.leecode.string
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* @Author: dingjiawen
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* @CreateTime: 2022-09-12 12:15
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* @Description: TODO 力扣459题:重复的子字符串:
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* 给定一个非空的字符串 s ，检查是否可以通过由它的一个子串重复多次构成。
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* @Version: 1.0
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*/
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public class RepeatedSubstringPattern {
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@Test
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public void test() {
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String s = "abcabcabc";
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System.out.println(repeatedSubstringPattern(s));
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// int[] next1 = getNext1(s);
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// System.out.println(Arrays.toString(next1));
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}
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/**
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* 这一题也可以用经典的KMP算法进行解决,又KMP算法可知:如果string由重复子字符串组成，其next数组一定是前面全是-1,后面全是升序排列
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* 如"abcabcabc",其next数组为[-1, -1, -1, 0, 1, 2, 3, 4, 5],所以规律是最后一个数+1可以整除是-1的数
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* 速度击败74.63%，内存击败10.85%
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* @param s
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* @return
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*/
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public boolean repeatedSubstringPattern(String s) {
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int[] next1 = getNext1(s);
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int length = next1.length;
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if (next1[length - 1] != -1 && length % (length - next1[length - 1] - 1) == 0) {
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return true;
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}
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return false;
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}
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public int[] getNext1(String needle) {
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int[] next = new int[needle.length()];
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next[0] = -1;
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int k = -1;
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for (int i = 1; i < needle.length(); ++i) {
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while (k != -1 && needle.charAt(k + 1) != needle.charAt(i)) {
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k = next[k];
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}
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if (needle.charAt(k + 1) == needle.charAt(i)) {
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++k;
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}
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next[i] = k;
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}
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return next;
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}
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}
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@ -0,0 +1,167 @@
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package com.markilue.leecode.string;
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import org.junit.Test;
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import java.util.Arrays;
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/**
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* @BelongsProject: Leecode
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* @BelongsPackage: com.markilue.leecode.string
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* @Author: dingjiawen
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* @CreateTime: 2022-09-12 09:01
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* @Description: TODO 力扣28题:实现strStr()
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* 实现strStr()函数。
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* 给你两个字符串haystack 和 needle ，请你在 haystack 字符串中找出 needle 字符串出现的第一个位置（下标从 0 开始）。如果不存在，则返回 -1 。
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* <p>
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* 说明：
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* 当needle是空字符串时，我们应当返回什么值呢？这是一个在面试中很好的问题。
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* 对于本题而言，当needle是空字符串时我们应当返回 0 。这与 C 语言的strstr()以及 Java 的indexOf()定义相符。
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* @Version: 1.0
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*/
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public class StrStr {
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@Test
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public void test() {
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String haystack = "hello";
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String needle = "ll";
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int i = strStr(haystack, needle);
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System.out.println(i);
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}
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@Test
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public void test1() {
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String haystack = "aaaaa";
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String needle = "bba";
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int i = strStr(haystack, needle);
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System.out.println(i);
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}
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@Test
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public void test2() {
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String haystack = "baabbbbababbbabab";
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String needle = "abbab";
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int i = strStr(haystack, needle);
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System.out.println(i);
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}
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@Test
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public void test3() {
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String haystack = "ababaabbbbababbaabaaabaabbaaaabbabaabbbbbbabbaabbabbbabbbbbaaabaababbbaabbbabbbaabbbbaaabbababbabbbabaaabbaabbabababbbaaaaaaababbabaababaabbbbaaabbbabb";
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String needle = "abbabbbabaa";
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int i = strStr(haystack, needle);
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System.out.println(i);
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}
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@Test
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public void test4() {
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String haystack = "ababac";
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// String needle = "abbabbbabaa";
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int[] next = getNext1(haystack);
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System.out.println(Arrays.toString(next));
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}
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/**
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* 尝试使用KMP算法:维护一个next数组,存储needle的最小前缀和最小后缀匹配的长度,从而减少从头开始匹配的次数,从而将时间复杂度从O(n*m)降到O(n+m)
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* 具体内容参考代码随想录和数据结构与算法之美的讲解内容
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* 速度击败39.11%，内存击败91.79%
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*
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* @param haystack
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* @param needle
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* @return
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*/
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public int strStr(String haystack, String needle) {
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int[] next = getNext(needle);
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int j = 0;
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for (int i = 0; i < haystack.length(); ++i) {
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while (j > 0 && haystack.charAt(i) != needle.charAt(j)) {
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j = next[j - 1] + 1;
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}
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if (haystack.charAt(i) == needle.charAt(j)) {
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++j;
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}
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if (j == needle.length()) {
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return i - j + 1;
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}
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}
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return -1;
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}
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public int[] getNext(String needle) {
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int[] result = new int[needle.length()];
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for (int i = 0; i < needle.length(); i++) {
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result[i] = -1;
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if (i == 0) {
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continue;
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}
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int right = i;
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for (int left = i - 1; left >= 0; left--) {
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boolean flag = false;
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if (needle.charAt(left) == needle.charAt(right)) {
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flag = true;
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for (int z = left - 1, k = right - 1; z >= 0; z--, k--) {
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if (needle.charAt(z) != needle.charAt(k)) {
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flag = false;
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break;
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}
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}
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}
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if (flag) {
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result[i] = left;
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break;
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}
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}
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}
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return result;
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}
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/**
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* 官方获取next数组的方法：精髓在于这次的next[i]可以通过上一次的next[i-1]来获取，分为两种情况：
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* 1）如果next[i-1]的下一个数和next[i]的下一个数相等，那么最大子串一定是在上一个的基础上+1,对应下面的if的内容
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* 2）如果next[i-1]的下一个数和next[i]的下一个数不相等，那么寻找上一个最大子串，看看上一次最大子串的下一个数和next[i]的下一个数相等，如果不相等就寻找在上一次的，对应于下面的while内容
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* 下所述的看就是用来记录上一次最大子串的索引
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*
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* 速度击败39.11%，内存击败73.88%
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* 但是只用了1ms
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* @param needle
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* @return
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*/
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public int[] getNext1(String needle) {
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int[] next = new int[needle.length()];
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next[0] = -1;
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int k = -1;
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for (int i = 1; i < needle.length(); ++i) {
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while (k != -1 && needle.charAt(k + 1) != needle.charAt(i)) {
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k = next[k];
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}
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if (needle.charAt(k + 1) == needle.charAt(i)) {
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++k;
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}
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next[i] = k;
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}
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return next;
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}
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}
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