leetcode240: 搜索二维矩阵 II
Opened this issue · 1 comments
wangyewei commented
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。
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
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 <= 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
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/search-a-2d-matrix-ii
wangyewei commented
思路和算法
二分查找
因为矩阵每行元素都是升序的,所以只需要对每行进行一次二分查找,判断target
是否出现过
/**
* @param {number[][]} matrix
* @param {number} target
* @return {boolean}
*/
var searchMatrix = function(matrix, target) {
const binary = (rows, target) => {
let low = 0
let high = rows.length - 1
while(low <= high) {
const mid = Math.floor((high - low) / 2) + low
const row = rows[mid]
if(row === target) {
return mid
}else if(row > target) {
high = mid - 1
}else {
low = mid + 1
}
}
return -1
}
for(const rows of matrix) {
if(binary(rows, target) != -1) return true
}
return false
};
复杂度分析
- 时间复杂度: O(m log n),每一行时间复杂度为
O(logn)
,一共m行 - 空间复杂度:O(1)