leetcode 240. Search a 2D Matrix II 矩阵搜索 + 右上角搜索

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.

对于排序好的矩阵，右上角开始搜索。

这道题很棒。

建议和leetcode 378. Kth Smallest Element in a Sorted Matrix 和 leetcode 668. Kth Smallest Number in Multiplication Table 有序矩阵搜索

代码如下：

/* * 右上角搜索 * */class Solution {    public boolean searchMatrix(int[][] matrix, int target)     {        if(matrix==null || matrix.length<=0)            return false;        int i=0 , j=matrix[0].length-1;        while(i<matrix.length && j>=0)        {            if(matrix[i][j]==target)                return true;            else if(matrix[i][j]>target)                j--;            else                 i++;        }        return false;    }}

下面是C++的做法，就是一个右上角搜索的做法，很经典

代码如下：

#include <iostream>#include <alogrithm>#include <set>#include <map>#include <vector>#include <stack>#include <queue>using namespace std;class Solution {public:    bool searchMatrix(vector<vector<int>>& mat, int target)     {        if (mat.size() <= 0)            return false;        int i = 0;        int j = mat[0].size() - 1;        while (i < mat.size() && j >= 0)        {            if (mat[i][j] == target)                return true;            else if (target < mat[i][j])                j--;            else                i++;        }        return false;    }};