【Leet Code】74. Search a 2D Matrix---Medium

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

.

思路：

解决该题目的方法就是对一维数组的二分查找进行变形操作，先对中间行对一维数组的二分查找，如果找到就返回true。

如果没找到，如果target<matrix[mid][0],则令down=mid-1；否则，up=mid+1.

代码实现：

class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
if(matrix.size() < 1 || matrix[0].size() < 1) return false;
int up = 0, down = matrix.size()-1;
int mid = 0;
while(up <= down)
{
mid = (up + down)/2;
if(searchArray(matrix[mid], target)) return true;

if(target < matrix[mid][0]) down = mid -1;
else up = mid + 1;
}
return false;
}
private:
bool searchArray(vector<int>& nums, int target){
int left = 0, right = nums.size() -1;
int mid = 0;
while(left <= right)
{
mid = (left + right)/2;
if(target == nums[mid])
return true;
if(target > nums[mid])
left = mid + 1;
else right = mid - 1;
}
return false;
}

};