comments	difficulty	edit_url	tags
true	中等	https://github.com/doocs/leetcode/edit/main/solution/0000-0099/0074.Search%20a%202D%20Matrix/README.md	数组 二分查找 矩阵

74. 搜索二维矩阵

English Version

题目描述

给你一个满足下述两条属性的 m x n 整数矩阵：

• 每行中的整数从左到右按非严格递增顺序排列。
• 每行的第一个整数大于前一行的最后一个整数。

给你一个整数 target ，如果 target 在矩阵中，返回 true ；否则，返回 false 。

 

示例 1：

输入：matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
输出：true

示例 2：

输入：matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13
输出：false

 

提示：

• m == matrix.length
• n == matrix[i].length
• 1 <= m, n <= 100
• -104 <= matrix[i][j], target <= 104

解法

方法一：二分查找

我们将二维矩阵逻辑展开，然后二分查找即可。

时间复杂度 $O(\log (m\times n))$。其中 $m$ 和 $n$ 分别是矩阵的行数和列数。空间复杂度 $O(1)$。

Python3

class Solution:
    def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
        m, n = len(matrix), len(matrix[0])
        left, right = 0, m * n - 1
        while left < right:
            mid = (left + right) >> 1
            x, y = divmod(mid, n)
            if matrix[x][y] >= target:
                right = mid
            else:
                left = mid + 1
        return matrix[left // n][left % n] == target

Java

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        int m = matrix.length, n = matrix[0].length;
        int left = 0, right = m * n - 1;
        while (left < right) {
            int mid = (left + right) >> 1;
            int x = mid / n, y = mid % n;
            if (matrix[x][y] >= target) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return matrix[left / n][left % n] == target;
    }
}

C++

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        int m = matrix.size(), n = matrix[0].size();
        int left = 0, right = m * n - 1;
        while (left < right) {
            int mid = left + right >> 1;
            int x = mid / n, y = mid % n;
            if (matrix[x][y] >= target) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return matrix[left / n][left % n] == target;
    }
};

Go

func searchMatrix(matrix [][]int, target int) bool {
	m, n := len(matrix), len(matrix[0])
	left, right := 0, m*n-1
	for left < right {
		mid := (left + right) >> 1
		x, y := mid/n, mid%n
		if matrix[x][y] >= target {
			right = mid
		} else {
			left = mid + 1
		}
	}
	return matrix[left/n][left%n] == target
}

TypeScript

function searchMatrix(matrix: number[][], target: number): boolean {
    const m = matrix.length;
    const n = matrix[0].length;
    let left = 0;
    let right = m * n;
    while (left < right) {
        const mid = (left + right) >>> 1;
        const i = Math.floor(mid / n);
        const j = mid % n;
        if (matrix[i][j] === target) {
            return true;
        }

        if (matrix[i][j] < target) {
            left = mid + 1;
        } else {
            right = mid;
        }
    }
    return false;
}

Rust

use std::cmp::Ordering;
impl Solution {
    pub fn search_matrix(matrix: Vec<Vec<i32>>, target: i32) -> bool {
        let m = matrix.len();
        let n = matrix[0].len();
        let mut i = 0;
        let mut j = n;
        while i < m && j > 0 {
            match matrix[i][j - 1].cmp(&target) {
                Ordering::Equal => {
                    return true;
                }
                Ordering::Less => {
                    i += 1;
                }
                Ordering::Greater => {
                    j -= 1;
                }
            }
        }
        false
    }
}

JavaScript

/**
 * @param {number[][]} matrix
 * @param {number} target
 * @return {boolean}
 */
var searchMatrix = function (matrix, target) {
    const m = matrix.length,
        n = matrix[0].length;
    let left = 0,
        right = m * n - 1;
    while (left < right) {
        const mid = (left + right + 1) >> 1;
        const x = Math.floor(mid / n);
        const y = mid % n;
        if (matrix[x][y] <= target) {
            left = mid;
        } else {
            right = mid - 1;
        }
    }
    return matrix[Math.floor(left / n)][left % n] == target;
};

方法二：从左下角或右上角搜索

这里我们以左下角作为起始搜索点，往右上方向开始搜索，比较当前元素 $matrix[i][j]$ 与 $target$ 的大小关系：

• 若 $matrix[i][j]=target$，说明找到了目标值，直接返回 true。
• 若 $matrix[i][j]>target$，说明这一行从当前位置开始往右的所有元素均大于 target，应该让 i 指针往上移动，即 $i=i-1$。
• 若 $matrix[i][j]<target$，说明这一列从当前位置开始往上的所有元素均小于 target，应该让 j 指针往右移动，即 $j=j+1$。

若搜索结束依然找不到 $target$，返回 false。

时间复杂度 $O(m+n)$。其中 $m$ 和 $n$ 分别是矩阵的行数和列数。空间复杂度 $O(1)$。

Python3

class Solution:
    def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
        m, n = len(matrix), len(matrix[0])
        i, j = m - 1, 0
        while i >= 0 and j < n:
            if matrix[i][j] == target:
                return True
            if matrix[i][j] > target:
                i -= 1
            else:
                j += 1
        return False

Java

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        int m = matrix.length, n = matrix[0].length;
        for (int i = m - 1, j = 0; i >= 0 && j < n;) {
            if (matrix[i][j] == target) {
                return true;
            }
            if (matrix[i][j] > target) {
                --i;
            } else {
                ++j;
            }
        }
        return false;
    }
}

C++

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        int m = matrix.size(), n = matrix[0].size();
        for (int i = m - 1, j = 0; i >= 0 && j < n;) {
            if (matrix[i][j] == target) return true;
            if (matrix[i][j] > target)
                --i;
            else
                ++j;
        }
        return false;
    }
};

Go

func searchMatrix(matrix [][]int, target int) bool {
	m, n := len(matrix), len(matrix[0])
	for i, j := m-1, 0; i >= 0 && j < n; {
		if matrix[i][j] == target {
			return true
		}
		if matrix[i][j] > target {
			i--
		} else {
			j++
		}
	}
	return false
}

Rust

use std::cmp::Ordering;
impl Solution {
    pub fn search_matrix(matrix: Vec<Vec<i32>>, target: i32) -> bool {
        let m = matrix.len();
        let n = matrix[0].len();
        let mut left = 0;
        let mut right = m * n;
        while left < right {
            let mid = left + (right - left) / 2;
            let i = mid / n;
            let j = mid % n;
            match matrix[i][j].cmp(&target) {
                Ordering::Equal => {
                    return true;
                }
                Ordering::Less => {
                    left = mid + 1;
                }
                Ordering::Greater => {
                    right = mid;
                }
            }
        }
        false
    }
}

JavaScript

/**
 * @param {number[][]} matrix
 * @param {number} target
 * @return {boolean}
 */
var searchMatrix = function (matrix, target) {
    const m = matrix.length,
        n = matrix[0].length;
    for (let i = m - 1, j = 0; i >= 0 && j < n; ) {
        if (matrix[i][j] == target) {
            return true;
        }
        if (matrix[i][j] > target) {
            --i;
        } else {
            ++j;
        }
    }
    return false;
};