// Source : https://oj.leetcode.com/problems/search-a-2d-matrix/ // Author : Hao Chen // Date : 2014-06-23 /********************************************************************************** * * 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. * **********************************************************************************/ class Solution { public: bool searchMatrix(vector>& matrix, int target) { return searchMatrix01(matrix, target); return searchMatrix02(matrix, target); } //Just simply convert the 2D matrix to 1D array. bool searchMatrix01(vector>& matrix, int target) { int row = matrix.size(); int col = row>0 ? matrix[0].size() : 0; int len = row * col; int low = 0, high = len -1; while (low <= high) { int mid = low + (high - low) / 2; int r = mid / col; int c = mid % col; int n = matrix[r][c]; if (n == target) return true; if (n < target) low = mid+1; else high = mid -1; } return false; } bool searchMatrix02(vector > &matrix, int target) { int idx = vertical_binary_search(matrix, target); if (idx<0){ return false; } idx = binary_search(matrix[idx], target); return (idx < 0 ? false : true); } int vertical_binary_search(vector< vector > v, int key){ int low = 0; int high = v.size()-1; while(low <= high){ int mid = low + (high-low)/2; if (v[mid][0] == key){ return mid; } if (key < v[mid][0]){ high = mid - 1; continue; } if (key > v[mid][0]){ low = mid + 1; continue; } } return low-1; } int binary_search(vector v, int key) { int low = 0; int high = v.size()-1; while(low <= high){ int mid = low + (high-low)/2; if (v[mid] == key){ return mid; } if (key < v[mid]){ high = mid - 1; continue; } if (key > v[mid]){ low = mid + 1; continue; } } return -1; } };