【题目】
Write an efficient algorithm that searches for a value in anmxnmatrix. This matrix has the following properties:
- Integers in each row are sorted from left to right.
- The first integer of each row is greater than the last integer of the previous row.
For example,
Consider the following matrix:
[
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]
Giventarget=3
, returntrue
.
【思路一】
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)
【代码】
/*------------------------------------
* 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;
}
【思路二】
根据排序规则,我们可以把矩阵看成一个串联起来的一维数组。正好可以利用二分查找。
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]
时间复杂度:O(nmlognm)
【代码二】
/*------------------------------------
* 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;
}
};
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