240\. 搜索二维矩阵 II
# 240. 搜索二维矩阵 II (opens new window)
# Description
Difficulty: 中等
Related Topics: 数组 (opens new window), 二分查找 (opens new window), 分治 (opens new window), 矩阵 (opens new window)
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
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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
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提示:
m == matrix.length
n == matrix[i].length
1 <= n, m <= 300
- -109 <= matrix[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
# Solution
Language: JavaScript
双指针
二分搜索
/**
* @param {number[][]} matrix
* @param {number} target
* @return {boolean}
*/
var searchMatrix = function(matrix, target) {
if(matrix.length === 0) return false
let row = matrix.length , col = matrix[0].length, i = row - 1, j = 0
while( i >= 0 && j < col ) {
if(matrix[i][j] > target) {
i--
} else if(matrix[i][j] < target) {
j++
} else {
return true
}
}
return false
};
var searchMatrix = function(matrix, target) {
if(matrix === null || matrix.length === 0 || matrix[0].length === 0) return false;
let col = 0;
let row = matrix[0].length - 1;
while(col < matrix.length && row >=0) {
if(matrix[col][row] > target) {
row--;
} else if(matrix[col][row] < target) {
col++
} else {
return true;
}
}
return false;
};
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