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240. 搜索二维矩阵 II

English Version

题目描述

编写一个高效的算法来搜索 m x n 矩阵 matrix 中的一个目标值 target 。该矩阵具有以下特性：

  • 每行的元素从左到右升序排列。
  • 每列的元素从上到下升序排列。

 

示例 1：

输入：matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 5
输出：true

示例 2：

输入：matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 20
输出：false

 

提示：

  • m == matrix.length
  • n == matrix[i].length
  • 1 <= n, m <= 300
  • -109 <= matix[i][j] <= 109
  • 每行的所有元素从左到右升序排列
  • 每列的所有元素从上到下升序排列
  • -109 <= target <= 109

解法

从左下角（或右上角）开始查找即可。

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

TypeScript

function searchMatrix(matrix: number[][], target: number): boolean {
    let m = matrix.length, n = matrix[0].length;
    let i = m - 1, j = 0;
    while (i >= 0 && j < n) {
        let cur = matrix[i][j];
        if (cur == target) return true;
        if (cur > 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();
        int i = m - 1, j = 0;
        while (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])
	i, j := m-1, 0
	for i >= 0 && j < n {
		if matrix[i][j] == target {
			return true
		}
		if matrix[i][j] > target {
			i--
		} else {
			j++
		}
	}
	return false
}

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