leecode更新

Leecode/src/main/java/com/markilue/leecode/array/MinSubArrayLen.java

@@ -0,0 +1,139 @@
+package com.markilue.leecode.array;
+
+import org.junit.Test;
+
+import javax.management.remote.rmi._RMIConnection_Stub;
+
+/**
+ * 寻找长度最小的子数组
+ */
+public class MinSubArrayLen {
+
+
+    @Test
+    public void test() {
+
+        int target = 3;
+        int[] nums = {1, 1, 1};
+        System.out.println(minSubArrayLen2(target, nums));
+
+    }
+
+
+    /**
+     * 代码随想录中解法:可伸缩的滑动窗口
+     * 看了思路之后自己尝试
+     * 速度击败20% 内存击败60%
+     * @param target
+     * @param nums
+     * @return
+     */
+    public int minSubArrayLen(int target, int[] nums) {
+
+        int start = 0;
+        int stop = 0;
+        //记录窗口总和
+        int sum = nums[start];
+        //记录窗口长度
+        int len = 0;
+        //记录结果
+        int result = nums.length;
+        boolean flag = false;
+        while (stop <= nums.length) {
+
+            while (sum >= target) {
+                flag = true;
+                len = stop - start + 1;
+                if (result > len) {
+                    result = len;
+                }
+                if (len == 1) {
+                    break;
+                }
+                sum -= nums[start++];
+            }
+            if (len == 1 && sum > target) {
+                return 1;
+            }
+            if (stop == nums.length - 1) {
+                break;
+            }
+
+            sum += nums[++stop];
+        }
+        if (flag) {
+            return result;
+        } else {
+            return 0;
+        }
+
+
+    }
+
+    /**
+     * 代码随想录中解法:可伸缩的滑动窗口
+     * 官方实现：速度击败99.99%，内存击败58.84%
+     *
+     * @param target
+     * @param nums
+     * @return
+     */
+    public int minSubArrayLen1(int target, int[] nums) {
+
+        int start = 0;
+        int sum = 0;
+        int result = Integer.MAX_VALUE;
+        int len = 0;
+
+        for (int stop = 0; stop < nums.length; stop++) {
+            sum += nums[stop];
+            while (sum >= target) {
+                len = stop - start + 1;
+                result = result > len ? len : result;
+                sum -= nums[start++];
+            }
+        }
+
+        return result == Integer.MAX_VALUE ? 0 : result;
+
+
+    }
+
+    /**
+     * 评论里给的滑动窗口法优化：
+     * 目前没看出优化在哪里，感觉没必要
+     * @param s
+     * @param nums
+     * @return
+     */
+    public int minSubArrayLen2(int s, int[] nums) {
+        if (nums.length == 0) {
+            return 0;
+        }
+        int count = 0;
+        int left = 0;
+        int right = 0;
+        while (count < s) {
+            if (right == nums.length) {
+                return 0;
+            }
+            count += nums[right];
+            right++;
+        }
+        while (right < nums.length) {
+            if (count < s) {
+                count += nums[right];
+                right++;
+            }
+            count -= nums[left];
+            left++;
+        }
+        while (count >= s) {
+            count -= nums[left];
+            left++;
+        }
+        return right - left + 1;
+    }
+
+
+}

Leecode/src/main/java/com/markilue/leecode/array/generateMatrix.java

@@ -0,0 +1,62 @@
+package com.markilue.leecode.array;
+
+import org.junit.Test;
+
+import java.util.Arrays;
+
+/**
+ * 螺旋矩阵:
+ * 给你一个正整数 n ，生成一个包含 1 到 n2 所有元素，且元素按顺时针顺序螺旋排列的 n x n 正方形矩阵 matrix
+ */
+public class generateMatrix {
+
+
+    @Test
+    public void test(){
+
+        int[][] ints = generateMatrix(6);
+        for (int[] anInt : ints) {
+            System.out.println(Arrays.toString(anInt));
+        }
+
+
+    }
+
+    /**
+     * 看了代码随想录之后的自己编程:寻找循环不变量
+     * 速度击败100%，内存击败87%
+     * 不会就先把第一层循环写出来，在找规律
+     * @param n
+     * @return
+     */
+    public int[][] generateMatrix(int n) {
+        int[][] result = new int[n][n];
+        int count = 1;
+        int i=0;
+        int j=0;
+        for (int k = 0; k < n / 2; k++) {
+
+            for (; i < result[k].length-k-1; i++) {
+                result[j][i]=count++;
+            }
+            for (; j < result[k].length-k-1; j++) {
+                result[j][i]=count++;
+            }
+            for (;i>k;i--){
+                result[j][i]=count++;
+            }
+            for (;j>k;j--){
+                result[j][i]=count++;
+            }
+            i+=1;
+            j+=1;
+
+        }
+        //如果是奇数,单独填充中间的
+        if(n%2==1){
+            result[n/2][n/2]=count++;
+        }
+        return result;
+
+    }
+}