Problem: 74. 搜索二维矩阵

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$m\times n$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:

$O(logmn)$

空间复杂度:

$O(mn)$

思路2：
时间复杂度:

$O(logmn)$

空间复杂度:

$O(1)$

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 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;
     }
 };