[LeetCode 240.] Search a 2D Matrix II

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