Leetcode:240. Search a 2D Matrix II (week 11)
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Description
Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:
Integers in each row are sorted in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.
For example,
Consider the following 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]
]
Given target = 5, return true.
Given target = 20, return false
解题思路
本题可用分治法解决, 以题目给出矩阵为例,查找数字5。仔细观察矩阵,最右上角的数字为15,由于矩阵是列递增,所以数字5不可能在最右侧15这一列,我们便可将这一列不予考虑,将范围缩减了一列。
[1, 4, 7, 11]
[2, 5, 8, 12]
[3, 6, 9, 16]
[10, 13, 14, 17]
[18, 21, 23, 26]
再判断数字11,同样
11>5
,又缩减一列。数字7同样小于5,在缩减一列,那么现在的矩阵变为:
[1, 4,]
[2, 5]
[3, 6]
[10, 13]
[18, 21]
判断数字4时,由于
5>4
,目标值肯定不在4所在的行,去点这一行,在进行判断。
[2, 5]
[3, 6]
[10, 13]
[18, 21]
判断数字5,找到目标值返回。 算法时间复杂度O(n)
代码
class Solution {public: bool searchMatrix(vector<vector<int>>& matrix, int target) { if(matrix.size() == 0) return false; int i = 0 , row = matrix.size(), col = matrix[0].size(); int j = col -1; while (i < row && j >= 0) { if (matrix[i][j] == target) return true; else if (matrix[i][j] > target) j--; else i++; } return false; }};
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