Problem: 74. 搜索二维矩阵

文章目录

  • 题目描述
  • 思路
  • 复杂度
  • Code

题目描述

思路

思路1:映射为一维数组二分查找

1.由于题目矩阵中的元素整体是升序的,我们可以将其放置在一个大小为m×nm \times n m×n的一维数组array中进行二分查找
2.对应的映射关系是array[mid] == mar[mid / n][mid % n]

思路2:直接在二维矩阵上进行二分查找

1.先对二维矩阵的第一列进行二分查找,找到小于等于target的一个数,讲此行标记为rowInd
2.从rowInd开始再进行二分查找

复杂度

思路1:
时间复杂度:

O(logmn)O(logmn) O(logmn)

空间复杂度:

O(mn)O(mn) O(mn)

思路2:
时间复杂度:

O(logmn)O(logmn) O(logmn)

空间复杂度:

O(1)O(1) 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 bounds if (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; } };