leecode，rt_phm更新更新

Leecode/src/main/java/com/markilue/leecode/hot100/second/T65_208_Trie_2.java

@@ -0,0 +1,53 @@
+package com.markilue.leecode.hot100.second;
+
+/**
+ *@BelongsProject: Leecode
+ *@BelongsPackage: com.markilue.leecode.hot100.second
+ *@Author: markilue
+ *@CreateTime: 2023-06-02  11:05
+ *@Description: TODO  力扣208  实现Trie(前缀树)
+ *@Version: 1.0
+ */
+public class T65_208_Trie_2 {
+
+    boolean isEnd;
+    T65_208_Trie_2[] children;
+
+    public T65_208_Trie_2() {
+        children = new T65_208_Trie_2[26];
+        isEnd = false;
+    }
+
+    public void insert(String word) {
+        T65_208_Trie_2 cur = this;
+        for (int i = 0; i < word.length(); i++) {
+            char c = word.charAt(i);
+            if (cur.children[c - 'a'] == null) {
+                cur.children[c - 'a'] = new T65_208_Trie_2();
+            }
+            cur = cur.children[c - 'a'];
+        }
+        cur.isEnd = true;//最优一个是终点
+    }
+
+    public boolean search(String word) {
+        T65_208_Trie_2 node = searchPredix(word);
+        return node != null && node.isEnd;
+    }
+
+    public boolean startsWith(String prefix) {
+        return searchPredix(prefix) != null;
+    }
+
+    public T65_208_Trie_2 searchPredix(String word) {
+        T65_208_Trie_2 cur = this;
+        for (int i = 0; i < word.length(); i++) {
+            char c = word.charAt(i);
+            if (cur.children[c - 'a'] == null) {
+                return null;//没有
+            }
+            cur = cur.children[c - 'a'];
+        }
+        return cur;//最优一个是终点
+    }
+}

Leecode/src/main/java/com/markilue/leecode/hot100/second/T81_312_MaxCoins.java

@@ -17,7 +17,7 @@ public class T81_312_MaxCoins {
     @Test
     public void test() {
         int[] nums = {3, 1, 5, 8};
-        System.out.println(maxCoins(nums));
+        System.out.println(maxCoins1(nums));
     }
 
 
@@ -62,4 +62,33 @@ public class T81_312_MaxCoins {
         }
         return memo[left][right];
     }
+
+
+    //由于戳气球会导致区间变成零散，所以可以考虑变成填气球；或者可以理解为一个区间最后一次戳的气球
+    public int maxCoins1(int[] nums) {
+        int n = nums.length;
+        int[] numNums = new int[n + 2];
+        for (int i = 1; i < n + 1; i++) {
+            numNums[i] = nums[i - 1];
+        }
+        numNums[0] = numNums[n + 1] = 1;
+        int[][] dp = new int[n + 2][n + 2];//dp[i][j]表示区间[i,j]的最大硬币数
+        for (int i = n - 1; i >= 0; i--) {
+            for (int j = i + 2; j <= n + 1; j++) {//区间至少大于3才能戳
+                for (int k = i + 1; k < j; k++) {
+                    //在左区间i和右区间j中间戳
+                    int sum = numNums[i] * numNums[k] * numNums[j];
+                    //加上左区间的和右区间的
+                    sum += dp[i][k] + dp[k][j];
+                    dp[i][j] = Math.max(dp[i][j], sum);
+                }
+            }
+        }
+
+        return dp[0][n + 1];
+
+
+    }
+
+
 }

Leecode/src/main/java/com/markilue/leecode/hot100/second/T86_394_DecodeString.java

@@ -23,6 +23,7 @@ public class T86_394_DecodeString {
     }
 
     int start = 0;
+
     public String sub(String s) {
         if (start >= s.length() || s.charAt(start) == ']') return "";
         StringBuilder result = new StringBuilder();
@@ -55,4 +56,37 @@ public class T86_394_DecodeString {
 
 
     }
+
+
+    public String subGet(String s) {
+        if (start > s.length() || s.charAt(start) == ']') return "";
+
+        char cur = s.charAt(start);
+        StringBuilder result = new StringBuilder();
+        if (Character.isDigit(cur)) {
+            int num = getNum(s);
+            //左括号
+            start++;
+            String next = subGet(s);
+            while (num-- > 0) {
+                result.append(next);
+            }
+            start++;//右括号
+        }else if(Character.isLetter(cur)){
+            result.append(cur);
+            start++;//下一个
+        }
+        //拼接剩下的
+        result.append(subGet(s));
+        return result.toString();
+    }
+
+    public int getNum(String s) {
+        int num = 0;
+        while (start < s.length() && Character.isDigit(s.charAt(start))) {
+            num = num * 10 + s.charAt(start) - '0';
+            start++;
+        }
+        return num;
+    }
 }

Leecode/src/main/java/com/markilue/leecode/interview/huawei/T0531/Question1.java

@@ -1,4 +1,4 @@
-package com.markilue.leecode.interview.huawei;
+package com.markilue.leecode.interview.huawei.T0531;
 
 import org.junit.Test;
 

Leecode/src/main/java/com/markilue/leecode/interview/huawei/T0531/Question2.java

@@ -1,4 +1,4 @@
-package com.markilue.leecode.interview.huawei;
+package com.markilue.leecode.interview.huawei.T0531;
 
 import org.junit.Test;
 

Leecode/src/main/java/com/markilue/leecode/interview/huawei/T0531/Question3.java

@@ -1,4 +1,4 @@
-package com.markilue.leecode.interview.huawei;
+package com.markilue.leecode.interview.huawei.T0531;
 
 import org.junit.Test;
 
@@ -37,7 +37,8 @@ public class Question3 {
                 {-1, 5, 2},
                 {-2, -3, 4},
         };
-        calculateMaxRectangleSum(2, 3, income);
+//        calculateMaxRectangleSum(2, 3, income);
+        calculate(2, 3, income);
     }
 
     private static void calculateMaxRectangleSum(int m, int n, int[][] matrix) {
@@ -70,4 +71,37 @@ public class Question3 {
     }
 
 
+    private static void calculate(int m, int n, int[][] matrix) {
+
+        //构造前缀和数组,计算前缀和
+        int[][] prefix = new int[m + 1][n + 1];
+
+        for (int i = 1; i < prefix.length; i++) {
+            for (int j = 1; j < prefix[0].length; j++) {
+                prefix[i][j] = prefix[i - 1][j] + prefix[i][j - 1] - prefix[i - 1][j - 1] + matrix[i - 1][j - 1];
+            }
+        }
+        int result = Integer.MIN_VALUE;
+        int edge = 0;
+        //利用前缀和来判断那个矩形最大
+        for (int i = 1; i < m + 1; i++) {
+            for (int j = 1; j < n + 1; j++) {
+                //左上角
+                for (int k = i; k < m + 1; k++) {
+                    for (int l = j; l < n + 1; l++) {
+                        //右下角
+                        int cur = prefix[k][l] - prefix[i - 1][l] - prefix[k][j - 1] + prefix[i - 1][j - 1];
+                        if (result < cur) {
+                            result = cur;
+                            edge = (k - i + 1) * (l - j + 1);
+                        }
+                    }
+                }
+            }
+        }
+        System.out.println(edge + " " + result);
+
+
+    }
+
 }