168 - 240 搜索二维矩阵2 - 001
题目
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:
现有矩阵 matrix 如下:
[ [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] ]
给定 target = 5,返回 true。
给定 target = 20,返回 false。
解答
从左下角开始,一点点向上找。如果遇到大的,就像上一行;遇到小的就向右一行。
class Solution:
def searchMatrix(self, matrix, target):
if len(matrix) == 0 or len(matrix[0]) == 0:
return False
height, width = len(matrix), len(matrix[0])
row, col = height-1, 0
while row >= 0 and col < width:
if matrix[row][col] > target:
row -= 1
elif matrix[row][col] < target:
col += 1
else:
return True
return False
Runtime: 36 ms, faster than 82.14% of Python3 online submissions for Search a 2D Matrix II.
Memory Usage: 17.4 MB, less than 92.59% of Python3 online submissions for Search a 2D Matrix II.
func searchMatrix(matrix [][]int, target int) bool {
if len(matrix) == 0 || len(matrix[0]) == 0 {
return false
}
height := len(matrix)
width := len(matrix[0])
row := height - 1
col := 0
for row >= 0 && col < width {
if matrix[row][col] > target {
row--
} else if matrix[row][col] < target {
col++
} else {
return true
}
}
return false
}
Runtime: 24 ms, faster than 89.76% of Go online submissions for Search a 2D Matrix II.
Memory Usage: 6.2 MB, less than 100.00% of Go online submissions for Search a 2D Matrix II.
var searchMatrix = function(matrix, target) {
if (matrix.length === 0 || matrix[0].length === 0) {
return false
}
let height = matrix.length
width = matrix[0].length
row = height - 1
col = 0
while (row >= 0 && col < width) {
if (matrix[row][col] > target) {
row--
} else if (matrix[row][col] < target) {
col++
} else {
return true
}
}
return false
};
Runtime: 80 ms, faster than 79.61% of JavaScript online submissions for Search a 2D Matrix II. Memory Usage: 37.5 MB, less than 33.33% of JavaScript online submissions for Search a 2D Matrix II.
Last updated
Was this helpful?