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.
这首题就纯用来练二分查找了。二分查找有时也可能会写错。注意保持区间一致。我习惯于区间左闭右闭。
1 class Solution { 2 public: 3 bool searchMatrix(vector<vector<int> > &matrix, int target) { 4 int row = matrix.size(); 5 if (row == 0) return false; 6 int col = matrix[0].size(); 7 if (col == 0) return false; 8 int mid, h = row - 1, l = 0; 9 while (h >= l) { 10 mid = (h + l) / 2; 11 if (target == matrix[mid][0] || target == matrix[mid][col - 1]) return true; 12 else if (target < matrix[mid][0]) { 13 h = mid - 1; 14 } else if (target > matrix[mid][col - 1]) { 15 l = mid + 1; 16 } else { 17 break; 18 } 19 } 20 if (h < l) return false; 21 h = col - 1, l = 0; 22 int m = 0; 23 while (h >= l) { 24 m = (h + l) / 2; 25 if (target == matrix[mid][m]) return true; 26 else if (target > matrix[mid][m]) { 27 l = m + 1; 28 } else { 29 h = m - 1; 30 } 31 } 32 return false; 33 } 34 };
LeetCode | Search a 2D Matrix,布布扣,bubuko.com
原文：http://www.cnblogs.com/linyx/p/3653881.html