[LeetCode] Search a 2D Matrix

Search a 2D Matrix

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 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]
]

Given target = 

3

, return 

true

.

解题思路：

题意为给定一个矩阵和一个目标值，判断目标值是否在矩阵中存在。矩阵满足：每一行从左往右递增，后一行的第一个元素大于前一行最后一个元素。

可以考虑二分查找法，将一维坐标转化成二维坐标即可。

class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
//二分查找，将矩阵查找转化成线性查找
int m = matrix.size();
if(m==0){
return false;
}
int n = matrix[0].size();
if(n==0){
return false;
}
int start = 0, end = m*n-1;
while(start<=end){
int middle = (start + end)/2;
int x = middle / n;
int y = middle % n;
if(matrix[x][y]==target){
return true;
}else if(matrix[x][y]<target){
start = middle + 1;
}else{
end = middle - 1;
}
}
return false;
}
};