A trickier binary search one.
////////////// class Solution { int lower_bound_vert(vector<vector<int>>& m, int col, int i, int j, int target) { int ret = j + 1; if (m[j][col] < target) return ret; else ret = j; int l = i, r = j; while (l < r) { int mid = l + ((r - l) >> 1); if (m[mid][col] == target) return mid; else if (m[mid][col] < target) l = mid + 1; else if (m[mid][col] > target) { r = mid; ret = mid; } } return ret; } public: bool searchMatrix(vector<vector<int>>& m, int target) { size_t rows = m.size(); if (rows == 0) return false; size_t cols = m[0].size(); // Prune vertically auto it0 = std::lower_bound(m[0].begin(), m[0].end(), target); if (it0 != m[0].end() && *it0 == target) return true; int max_c = 0; if (it0 == m[0].begin()) return false; else max_c = it0 - m[0].begin(); // Prune horizontally int max_r = lower_bound_vert(m, 0/*col_0*/, 0, rows - 1, target); if (max_r == 0) return false; if (max_r < rows && m[max_r][0] == target) return true; // Pick cols int start_row = lower_bound_vert(m, max_c - 1, 0, max_r - 1, target); if (start_row == max_r) return false; if (m[start_row][max_c - 1] == target) return true; while (start_row < max_r) { if (std::binary_search(m[start_row].begin(), m[start_row].begin() + max_c, target)) return true; start_row++; } return false; } };
But there are always smarter\shorter solution: https://leetcode.com/discuss/47528/c-with-o-m-n-complexity
