代码随想录算法训练营第二天| 977.有序数组的平方、 209.长度最小的子数组、 59.螺旋矩阵II

977.有序数组的平方
https://leetcode.cn/problems/squares-of-a-sorted-array/description/

public static int[] sortedSquares(int[] nums){
        int left = 0;
        int right = nums.length - 1;
        int[] result = new int[nums.length];
        int write = result.length - 1;
        while (left <= right){
            if (Math.pow(nums[left],2) > Math.pow(nums[right],2)) {
                result[write] = nums[left] * nums[left];
                left++;
                write--;
            }else {
                result[write] = nums[right] * nums[right];
                right--;
                write--;
            }
        }
        return result;
    }

总结：双指针每次在最大的两个之中去读，把大的添加到新数组中

** 209.长度最小的子数组**
https://leetcode.cn/problems/minimum-size-subarray-sum/description/

public static int minSubArrayLen(int target, int[] nums) {
        int left = 0;
        int sum = 0;
        int result = Integer.MAX_VALUE;
        for (int right = 0; right < nums.length; right++) {
            sum += nums[right];
            while (sum >= target){
                result = Math.min(result,right - left +1);
                sum -= nums[left++];
            }
        }
        return result == Integer.MAX_VALUE ? 0 : result;
    }

总结：滑动窗口的思想，right仅需一次遍历数组即可，right一直向前，无需回头
59.螺旋矩阵II
https://leetcode.cn/problems/spiral-matrix-ii/submissions/508708830/

public static int[][] generateMatrix(int n) {
        int l = 0,t = 0,r = n - 1,b = n - 1;
        int num = 1 , tar = n * n;
        int[][] result = new int[n][n];
        while (num <= tar){
            for (int i = l; i <= r; i++) {
                result[t][i] = num++;
            }
            t++;
            for (int i = t; i <= b; i++){
                result[i][r] = num++;
            }
            r--;
            for (int i = r; i >= l; i--){
                result[b][i] = num++;
            }
            b--;
            for (int i = b; i >= t; i--){
                result[i][l] = num++;
            }
            l++;
        }
        return result;
    }

总结：这种矩阵的题 关键就在于 上下左右四个边界定义好，定义好之后边界就包含了遍历的下标的信息，之后边界一直在变，就往里填数字就好，关键就在于边界！！！