程式碼隨想錄演算法訓練營第二天| 977.有序陣列的平方、 209.長度最小的子陣列、 59.螺旋矩陣II

977.有序陣列的平方
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;
    }

總結：雙指標每次在最大的兩個之中去讀，把大的新增到新陣列中

** 209.長度最小的子陣列**
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;
    }

總結：滑動視窗的思想，right僅需一次遍歷陣列即可，right一直向前，無需回頭
59.螺旋矩陣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;
    }

總結：這種矩陣的題 關鍵就在於 上下左右四個邊界定義好，定義好之後邊界就包含了遍歷的下標的資訊，之後邊界一直在變，就往裡填數字就好，關鍵就在於邊界！！！