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,递归。
class Solution { public: int m = 0, n = 0; vector<vector<int>> data; bool search(int left, int right, int target) { int mid = (right + left) / 2; if(right - left == 1) return target == data[left / n][left % n]; else return search(left, mid, target) || search(mid, right, target); } bool searchMatrix(vector<vector<int> > &matrix, int target) { // IMPORTANT: Please reset any member data you declared, as // the same Solution instance will be reused for each test case. if(matrix.size() == 0) return false; data = matrix; m = matrix.size(), n = matrix[0].size(); return search(0, m * n, target); } };