[LeetCode]Search a 2D Matrix

题目描述：(链接)

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：

二分查找

class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
int first = 0;
int last = matrix.size() - 1;
int len = matrix[0].size();
while (first <= last) {
int mid = first + (last - first) / 2;
if (matrix[mid][0] <= target && target <= matrix[mid][len - 1] ) {
return binarySearch(matrix[mid], target);
} else if (matrix[mid][0] > target) {
last = mid - 1;
} else if (matrix[mid][len - 1] < target) {
first = mid + 1;
}
}

return false;
}

private:
bool binarySearch(vector<int>& vec, int target) {
int first = 0;
int last = vec.size() - 1;
while (first <= last) {
int mid= first + (last -first) / 2;
if (vec[mid] == target) {
return true;
} else if (vec[mid] < target) {
first = mid + 1;
} else {
last = mid - 1;
}
}

return false;
}
};