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1351. 统计有序矩阵中的负数

English Version

题目描述

给你一个 m * n 的矩阵 grid，矩阵中的元素无论是按行还是按列，都以非递增顺序排列。 请你统计并返回 grid 中 负数 的数目。

 

示例 1：

输入：grid = [[4,3,2,-1],[3,2,1,-1],[1,1,-1,-2],[-1,-1,-2,-3]]
输出：8
解释：矩阵中共有 8 个负数。

示例 2：

输入：grid = [[3,2],[1,0]]
输出：0

 

提示：

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

 

进阶：你可以设计一个时间复杂度为 O(n + m) 的解决方案吗？

解法

方法一：从左下角或右上角开始遍历

从左下角开始往右上方向遍历。当遇到负数时，说明这一行从当前位置开始往右的所有元素均为负数，那么 ans += n - j，并且 i 指针上移；否则 j 指针右移。遍历结束，返回 ans。

时间复杂度 O(m + n)。

方法二：二分查找

遍历每一行，查找每一行第一个小于 0 的位置，从该位置开始往右的所有元素均为负数，累加负数个数到 ans。遍历结束，返回 ans。

时间复杂度 O(mlogn)。

Python3

class Solution:
    def countNegatives(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        i, j = m - 1, 0
        ans = 0
        while i >= 0 and j < n:
            if grid[i][j] < 0:
                ans += n - j
                i -= 1
            else:
                j += 1
        return ans

class Solution:
    def countNegatives(self, grid: List[List[int]]) -> int:
        ans = 0
        n = len(grid[0])
        for row in grid:
            left, right = 0, n
            while left < right:
                mid = (left + right) >> 1
                if row[mid] < 0:
                    right = mid
                else:
                    left = mid + 1
            ans += n - left
        return ans

Java

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

class Solution {
    public int countNegatives(int[][] grid) {
        int ans = 0;
        int n = grid[0].length;
        for (int[] row : grid) {
            int left = 0, right = n;
            while (left < right) {
                int mid = (left + right) >> 1;
                if (row[mid] < 0) {
                    right = mid;
                } else {
                    left = mid + 1;
                }
            }
            ans += n - left;
        }
        return ans;
    }
}

TypeScript

function countNegatives(grid: number[][]): number {
    const m = grid.length,
        n = grid[0].length;
    let ans = 0;
    for (let i = m - 1, j = 0; i >= 0 && j < n; ) {
        if (grid[i][j] < 0) {
            ans += n - j;
            --i;
        } else {
            ++j;
        }
    }
    return ans;
}

function countNegatives(grid: number[][]): number {
    const n = grid[0].length;
    let ans = 0;
    for (let row of grid) {
        let left = 0,
            right = n;
        while (left < right) {
            const mid = (left + right) >> 1;
            if (row[mid] < 0) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        ans += n - left;
    }
    return ans;
}

C++

class Solution {
public:
    int countNegatives(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        int ans = 0;
        for (int i = m - 1, j = 0; i >= 0 && j < n;)
        {
            if (grid[i][j] < 0)
            {
                ans += n - j;
                --i;
            }
            else ++j;
        }
        return ans;
    }
};

class Solution {
public:
    int countNegatives(vector<vector<int>>& grid) {
        int ans = 0;
        int n = grid[0].size();
        for (auto& row : grid)
        {
            int left = 0, right = n;
            while (left < right)
            {
                int mid = (left + right) >> 1;
                if (row[mid] < 0) right = mid;
                else left = mid + 1;
            }
            ans += n - left;
        }
        return ans;
    }
};

Go

func countNegatives(grid [][]int) int {
	m, n := len(grid), len(grid[0])
	ans := 0
	for i, j := m-1, 0; i >= 0 && j < n; {
		if grid[i][j] < 0 {
			ans += n - j
			i--
		} else {
			j++
		}
	}
	return ans
}

func countNegatives(grid [][]int) int {
	ans, n := 0, len(grid[0])
	for _, row := range grid {
		left, right := 0, n
		for left < right {
			mid := (left + right) >> 1
			if row[mid] < 0 {
				right = mid
			} else {
				left = mid + 1
			}
		}
		ans += n - left
	}
	return ans
}

JavaScript

/**
 * @param {number[][]} grid
 * @return {number}
 */
var countNegatives = function (grid) {
    const m = grid.length,
        n = grid[0].length;
    let ans = 0;
    for (let i = m - 1, j = 0; i >= 0 && j < n; ) {
        if (grid[i][j] < 0) {
            ans += n - j;
            --i;
        } else {
            ++j;
        }
    }
    return ans;
};

/**
 * @param {number[][]} grid
 * @return {number}
 */
var countNegatives = function (grid) {
    const n = grid[0].length;
    let ans = 0;
    for (let row of grid) {
        let left = 0,
            right = n;
        while (left < right) {
            const mid = (left + right) >> 1;
            if (row[mid] < 0) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        ans += n - left;
    }
    return ans;
};

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