Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:
For example,
Consider the following matrix:
[ [1, 3, 5, 7], [10, 11, 16, 20], [23, 30, 34, 50] ]
Given target = 3, return true.
杨氏矩阵解法：
（1）如果数组中选取的数字刚好和查找的数字相等，查找结束，返回true
（2）如果数组中选取的数字小于查找的数字，根据排序规则，要查找的数字只能在当前选取的数字的右边。列数减一。
（3）如果数组中选取的数字大于查找的数字，根据排序规则，要查找的数字只能在当前选取的数字的下边。行数加一。
时间复杂度：O(m+n)
/*------------------------------------
* 日期：2015-02-04
* 作者：SJF0115
* 题目: 74.Search a 2D Matrix
* 网址：https://oj.leetcode.com/problems/search-a-2d-matrix/
* 结果：AC
* 来源：LeetCode
* 博客：
---------------------------------------*/
#include <iostream>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;
class Solution {
public:
bool searchMatrix(vector<vector<int> > &matrix, int target) {
if(matrix.empty()){
return false;
}//if
int row = matrix.size();
int col = matrix[0].size();
int x = 0,y = col - 1;
// 杨氏矩阵解法从右上角元素开始
while(x < row && y >= 0){
if(matrix[x][y] == target){
return true;
}//if
else if(matrix[x][y] < target){
++x;
}//else
else{
--y;
}
}//while
return false;
}
};
int main(){
Solution s;
int target = 3;
vector<vector<int> > matrix = {{1,3,5,7},{10,11,16,20},{23,30,34,50}};
bool result = s.searchMatrix(matrix,target);
// 输出
cout<<result<<endl;
return 0;
}
n*m矩阵转换为一维数组：matrix[x][y] -> a[x*m + y]
一维数组转换为矩阵：a[x] -> matrix[x / m][ x % m] /*------------------------------------
* 日期：2015-02-04
* 作者：SJF0115
* 题目: 74.Search a 2D Matrix
* 网址：https://oj.leetcode.com/problems/search-a-2d-matrix/
* 结果：AC
* 来源：LeetCode
* 博客：
---------------------------------------*/
class Solution {
public:
bool searchMatrix(vector<vector<int> > &matrix, int target) {
if(matrix.empty()){
return false;
}//if
int row = matrix.size();
int col = matrix[0].size();
int start = 0,end = row * col - 1;
// 二分查找解法
int mid,x,y;
while(start <= end){
mid = start + (end - start) / 2;
x = mid / col;
y = mid % col;
if(matrix[x][y] == target){
return true;
}//if
else if(target > matrix[x][y]){
start = mid + 1;
}//else
else{
end = mid - 1;
}//else
}//while
return false;
}
};
[LeetCode]74.Search a 2D Matrix
原文地址：http://blog.csdn.net/sunnyyoona/article/details/43486229
