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Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:
For example,
Consider the following matrix:
[ [1, 3, 5, 7], [10, 11, 16, 20], [23, 30, 34, 50] ]
Given target = 3, return true.
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首先二分查找每行的第一个元素，确定了行号后，再在当前行内进行二分查找
值得一提的是其时间复杂度为 O(log(m) + log(n)) = O(log(m*n))
bool searchMatrix(vector<vector<int>>& matrix, int target) { int beg = 0, mid, end = matrix.size()-1; while (beg <= end) { mid = (beg + end) >> 1; if (target < matrix[mid][0]) end = mid - 1; else if (target > matrix[mid][0]) beg = mid + 1; else return true; } int row = end; if (row < 0) return false; beg = 0; end = matrix[row].size(); while (beg <= end) { mid = (beg + end) >> 1; if (target < matrix[row][mid]) end = mid - 1; else if (target > matrix[row][mid]) beg = mid + 1; else return true; } return false; }
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原文地址：http://www.cnblogs.com/sdlwlxf/p/5107588.html
