LeetCode 74. Search a 2D Matrix
2016-11-20 11:31
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描述
在一个有序矩阵里面找一个数解决
因为矩阵每行的行首大于上一行的末尾,可以当成一维的数组来二分搜索即可class Solution { public: bool searchMatrix(const vector<vector<int>>& matrix, int target) { int r = matrix.size(); if (!r) return false; int c = r ? matrix[0].size() : 0; int st = 0, ed = r * c - 1; while (st < ed) { int m = st + ((ed - st) >> 1); // cout << m << endl; int x = m / c, y = m % c; if (matrix[x][y] == target) { return true; } else if (matrix[x][y] < target) st = m + 1; else ed = m - 1; } if (matrix[st / c][st % c] == target) return true; return false; } };
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