Problem: 74. 搜索二維矩陣

題目描述

思路

思路1：映射為一維數組二分查找

1.由于題目矩陣中的元素整體是升序的，我們可以將其放置在一個大小為$m\times n$的一維數組array中進行二分查找
2.對應的映射關系是array[mid] == mar[mid / n][mid % n]

思路2：直接在二維矩陣上進行二分查找

1.先對二維矩陣的第一列進行二分查找，找到小于等于target的一個數,講此行標記為rowInd
2.從rowInd開始再進行二分查找

復雜度

思路1：
時間復雜度:

$O(logmn)$

空間復雜度:

$O(mn)$

思路2：
時間復雜度:

$O(logmn)$

空間復雜度:

$O(1)$

Code

思路1：

class Solution {
public:/*** Binary Search* * @param matrix Given array* @param target Given target number* @return bool*/bool searchMatrix(vector<vector<int>> &matrix, int target) {int row = matrix.size();if (row == 0) {return false;}int col = matrix[0].size();int left = 0;int right = row * col - 1;vector<int> array(row * col);while (left <= right) {int mid = left + (right - left) / 2;if (matrix[mid / col][mid % col] == target) {return true;} else if (matrix[mid / col][mid % col] > target) {right = mid - 1;} else if (matrix[mid / col][mid % col] < target) {left = mid + 1;}}return false;}
};

思路2：

class Solution {public:/*** Binary Search** @param matrix Given array* @param target Given target number* @return bool*/bool searchMatrix(vector<vector<int>>& matrix, int target) {int row = matrix.size();if (row == 0) {return false;}int col = matrix[0].size();int left = 0;int right = row - 1;while (left <= right) {int mid = left + (right - left) / 2;if (matrix[mid][0] == target) {return true;} else if (matrix[mid][0] > target) {right = mid - 1;} else if (matrix[mid][0] < target) {left = mid + 1;}}// Check out of boundsif (right < 0) {return false;}int rowIdx = right;left = 0;right = col - 1;while (left <= right) {int mid = left + (right - left) / 2;if (matrix[rowIdx][mid] == target) {return true;} else if (matrix[rowIdx][mid] > target) {right = mid - 1;} else if (matrix[rowIdx][mid] < target) {left = mid + 1;}}return false;}};