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
.
二分查找。思路简单,先定位在哪个数组中,再继续在数组中查找。定位在哪个数组中考察数组第一个值,实现类似lower_bound()就能定位在哪个数组中。[LeetCode]—Search Insert Position 有序数组中找目标插入的位置
class Solution {
public:
bool searchMatrix(vector<vector<int> > &matrix, int target) {
int n=matrix.size();
int first=0, last=n;
//定位那个数组中
while(first<last){
int mid=(first+last)/2;
if(matrix[mid][0]>target){
last=mid;
}else if(matrix[mid][0]<target){
first=mid+1;
}else
return true;
}
//last 返回的是类似lower_bound()返回的,应该插入的位置
if(last==0)return false;
last=last-1; //继续考察matrix[n-1]数组部分
//在matrix[last]数组中继续二分查找
return binary_search(matrix[last].begin(),matrix[last].end(),target);
}
};