LeetCode 240. Search a 2D Matrix II
一道经典的二维矩阵搜索题。
题目描述
Write an efficient algorithm that searches for a target value in an m x n integer matrix. The 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.
Example 1:
Input: 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
Output: true
Example 2:
Input: 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
Output: false
Constraints:
- m == matrix.length
- n == matrix[i].length
- 1 <= n, m <= 300
- -109 <= matix[i][j] <= 109
- All the integers in each row are sorted in ascending order.
- All the integers in each column are sorted in ascending order.
- -109 <= target <= 109
解题思路
这道题是典型的减治算法，关键在于这个中间点的选取。
本题可以选择副对角线上的两个顶点。只要选取的点与target不相等，便可以一次减去一行或者一列的查找，缩小搜索空间。
参考代码
/*
* @lc app=leetcode id=240 lang=cpp
*
* [240] Search a 2D Matrix II
*/
// @lc code=start
class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
assert(!matrix.empty() && !matrix[0].empty());
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) j++;
else i--;
}
return false;
}
};
// @lc code=end