1351. Count Negative Numbers in a Sorted Matrix

Given a m * n matrix grid which is sorted in non-increasing order both row-wise and column-wise.

Return the number of negative numbers in grid.

Example 1:

Input: grid = [[4,3,2,-1],[3,2,1,-1],[1,1,-1,-2],[-1,-1,-2,-3]]
Output: 8
Explanation: There are 8 negatives number in the matrix.

Example 2:

Input: grid = [[3,2],[1,0]]
Output: 0

Example 3:

Input: grid = [[1,-1],[-1,-1]]
Output: 3

Example 4:

Input: grid = [[-1]]
Output: 1

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 100
• -100 <= grid[i][j] <= 100

class Solution {
    public int countNegatives(int[][] grid) {
        int res = 0;
        for(int i = 0; i < grid.length; i++){
            for(int j = 0; j < grid[0].length; j++){
                if(grid[i][j] < 0) res++;
            }
        }
        return res;
    }
}

我没看错吧？brute force还66.9%

class Solution {
    public int countNegatives(int[][] grid) {
        int res = 0;
        int m = grid.length;
        int n = grid[0].length;
        int i = 0;
        int j = n - 1;
        while(i < m && j >= 0){
            if(grid[i][j] < 0){
                res += m - i;
                j--;
            }
            else i++;
        }
        return res;
    }
}

O(M + N)，从右上角开始扫到左下角

原文地址：https://www.cnblogs.com/wentiliangkaihua/p/12325484.html