240. Search a 2D Matrix II

题目：

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.

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.

翻译：

编写一个搜索m×n矩阵中的值的有效算法。 该矩阵具有以下属性：


每行中的整数按从左到右的顺序进行排序。
每列中的整数按照从上到下的顺序进行排序。
例如，


考虑以下矩阵：


[
   [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。


给定目标= 20，返回false。

解题代码：

#include<iostream>#include<vector>using namespace std;#include<iostream>#include<vector>//#include<> using namespace std;class Solution {public:    bool searchMatrix(vector<vector<int> >& matrix, int target) {    int m = matrix.size();    if(m == 0) return false;    int n = matrix[0].size();    int i = 0, j = n -1;    while(i < m && j >= 0) {    if(matrix[i][j] == target) return true;    else if(matrix[i][j] > target) {    j--;} else i++;}return false;    }};int main() {int arr[4][5] = {{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}  };vector<vector<int> > v;for(int i = 0; i < 4 ;i++) {v.push_back(vector<int> ( arr[i],arr[i]+5) );} Solution s; cout << s.searchMatrix(v,5) << " " << s.searchMatrix(v,20)<<endl;}

题目状态：