109 lines
2.8 KiB
C++
109 lines
2.8 KiB
C++
// Source : https://oj.leetcode.com/problems/search-a-2d-matrix/
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// Author : Hao Chen
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// Date : 2014-06-23
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/**********************************************************************************
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*
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* Write an efficient algorithm that searches for a value in an m x n matrix.
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* This matrix has the following properties:
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*
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* Integers in each row are sorted from left to right.
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* The first integer of each row is greater than the last integer of the previous row.
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*
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* For example,
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*
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* Consider the following matrix:
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*
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* [
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* [1, 3, 5, 7],
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* [10, 11, 16, 20],
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* [23, 30, 34, 50]
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* ]
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*
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* Given target = 3, return true.
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*
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**********************************************************************************/
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class Solution {
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public:
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bool searchMatrix(vector<vector<int>>& matrix, int target) {
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return searchMatrix01(matrix, target);
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return searchMatrix02(matrix, target);
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}
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//Just simply convert the 2D matrix to 1D array.
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bool searchMatrix01(vector<vector<int>>& matrix, int target) {
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int row = matrix.size();
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int col = row>0 ? matrix[0].size() : 0;
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int len = row * col;
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int low = 0, high = len -1;
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while (low <= high) {
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int mid = low + (high - low) / 2;
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int r = mid / col;
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int c = mid % col;
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int n = matrix[r][c];
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if (n == target) return true;
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if (n < target) low = mid+1;
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else high = mid -1;
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}
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return false;
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}
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bool searchMatrix02(vector<vector<int> > &matrix, int target) {
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int idx = vertical_binary_search(matrix, target);
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if (idx<0){
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return false;
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}
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idx = binary_search(matrix[idx], target);
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return (idx < 0 ? false : true);
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}
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int vertical_binary_search(vector< vector<int> > v, int key){
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int low = 0;
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int high = v.size()-1;
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while(low <= high){
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int mid = low + (high-low)/2;
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if (v[mid][0] == key){
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return mid;
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}
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if (key < v[mid][0]){
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high = mid - 1;
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continue;
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}
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if (key > v[mid][0]){
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low = mid + 1;
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continue;
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}
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}
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return low-1;
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}
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int binary_search(vector<int> v, int key) {
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int low = 0;
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int high = v.size()-1;
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while(low <= high){
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int mid = low + (high-low)/2;
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if (v[mid] == key){
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return mid;
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}
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if (key < v[mid]){
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high = mid - 1;
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continue;
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}
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if (key > v[mid]){
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low = mid + 1;
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continue;
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}
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}
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return -1;
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}
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};
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