LeetCodeHot100 73. 矩阵置零 54. 螺旋矩阵 48. 旋转图像 240. 搜索二维矩阵 II

73. 矩阵置零
https://leetcode.cn/problems/set-matrix-zeroes/description/?envType=study-plan-v2&envId=top-100-liked

public void setZeroes(int[][] matrix) {
        int top = 0,bottom = matrix.length,left = 0,right = matrix[0].length;
        int[][] flag = new int[bottom][right];
        for (int i = 0; i < bottom; i++) {
            for (int j = 0; j < right; j++) {
                if (matrix[i][j] == 0) flag[i][j] = 1;
            }
        }
        for (int i = 0; i < bottom; i++) {
            for (int j = 0; j < right; j++) {
                if (flag[i][j] == 1){
                    for (int k = 0; k < i; k++) {
                        matrix[k][j] = 0;
                    }
                    for (int k = i; k < bottom; k++) {
                        matrix[k][j] = 0;
                    }
                    for (int k = 0; k < j; k++) {
                        matrix[i][k] = 0;
                    }
                    for (int k = j; k < right; k++) {
                        matrix[i][k] = 0;
                    }
                }
            }
        }
    }

总结：new一个数组去记录哪个位置是0 ，再把0的位置去上下左右变成0
54. 螺旋矩阵
https://leetcode.cn/problems/spiral-matrix/description/?envType=study-plan-v2&envId=top-100-liked

public List<Integer> spiralOrder(int[][] matrix) {
        int top = 0 , bottom = matrix.length - 1 , left = 0 , right = matrix[0].length - 1;
        int num = 0 ;
        int tar = (bottom + 1) * (right + 1);
        List<Integer> list = new ArrayList<>();
        while (num < tar){
            for (int i = left; i <= right; i++) {
                list.add(matrix[top][i]);
                num++;
            }
            if (num == tar) break;;
            top++;
            for (int i = top; i <= bottom; i++) {
                list.add(matrix[i][right]);
                num++;
            }
            if (num == tar) break;;
            right--;
            for (int i = right; i >= left; i--) {
                list.add(matrix[bottom][i]);
                num++;
            }
            if (num == tar) break;;
            bottom--;
            for (int i = bottom; i >= top; i--) {
                list.add(matrix[i][left]);
                num++;
            }
            if (num == tar) break;;
            left++;
        }
        return list;
    }

总结：经典的矩阵问题，把top，left，bottom，right 这四个边界弄明白就没问题
48. 旋转图像
https://leetcode.cn/problems/rotate-image/description/?envType=study-plan-v2&envId=top-100-liked

public void rotate(int[][] matrix) {
        int top = 0,bottom = matrix.length - 1,left = 0,right = matrix[0].length - 1;
        //f用来标记选择的格子
        int n = matrix.length;
        for (int i = 0;i < n / 2;i++){
            for (int f = left; f < right; f++) {
                int iTop = matrix[top][f];
                int iRight = matrix[f][right];
                int iBottom = matrix[bottom][right - (f - left)];
                int iLeft = matrix[bottom - (f - left)][left];
                matrix[top][f] = iLeft;
                matrix[f][right] = iTop;
                matrix[bottom][right - (f - left)] = iRight;
                matrix[bottom - (f - left)][left] = iBottom;
            }
            top++;
            right--;
            bottom--;
            left++;
        }
    }

总结：一次转四个，yeap
240. 搜索二维矩阵 II
https://leetcode.cn/problems/search-a-2d-matrix-ii/?envType=study-plan-v2&envId=top-100-liked

public boolean searchMatrix(int[][] matrix, int target) {
        int m = matrix.length, n = matrix[0].length;
        int r = 0;
        int c = n - 1;
        while (r < m && c >= 0){
            if (matrix[r][c] == target) {
                return true;
            }else if (target > matrix[r][c]){
                r++;
            }else {
                c--;
            }
        }
        return false;
    }

总结：此题可以看成右上角为root的二叉排序树，很nice的思路，，无敌！