[LeetCode] 240. Search a 2D Matrix II

[grandyang]

 

Write an efficient algorithm that searches for a target value in an m x n integer matrix. The matrix has the following properties:

• Integers in each row are sorted in ascending from left to right.
• Integers in each column are sorted in ascending from top to bottom.

 

Example 1:

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 5
Output: true

Example 2:

Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 20
Output: false

 

Constraints:

• m == matrix.length
• n == matrix[i].length
• 1 <= n, m <= 300
• -109 <= matix[i][j] <= 109
• All the integers in each row are sorted in ascending order.
• All the integers in each column are sorted in ascending order.
• -109 <= target <= 109

 

突然发现 LeetCode 很喜欢从 LintCode 上盗题，这是逼我去刷 LintCode 的节奏么?! 这道题让我们在一个二维数组中快速的搜索的一个数字，这个二维数组各行各列都是按递增顺序排列的，是之前那道 Search a 2D Matrix 的延伸，那道题的不同在于每行的第一个数字比上一行的最后一个数字大，是一个整体蛇形递增的数组。所以那道题可以将二维数组展开成一个一位数组用一次二查搜索。而这道题没法那么做，这道题有它自己的特点。如果我们观察题目中给的那个例子，可以发现有两个位置的数字很有特点，左下角和右上角的数。左下角的 18，往上所有的数变小，往右所有数增加，那么就可以和目标数相比较，如果目标数大，就往右搜，如果目标数小，就往上搜。这样就可以判断目标数是否存在。当然也可以把起始数放在右上角，往左和下搜，停止条件设置正确就行。代码如下：

 

class Solution {
public:
    bool searchMatrix(vector<vector<int> > &matrix, int target) {
        if (matrix.empty() || matrix[0].empty()) return false;
        if (target < matrix[0][0] || target > matrix.back().back()) return false;
        int x = matrix.size() - 1, y = 0;
        while (true) {
            if (matrix[x][y] > target) --x;
            else if (matrix[x][y] < target) ++y;
            else return true;
            if (x < 0 || y >= matrix[0].size()) break;
        }
        return false;
    }
};

 

Github 同步地址：

#240

 

类似题目：

Search a 2D Matrix

 

参考资料：

https://leetcode.com/problems/search-a-2d-matrix-ii/

https://leetcode.com/problems/search-a-2d-matrix-ii/discuss/66139/C%2B%2B-search-from-top-right

https://leetcode.com/problems/search-a-2d-matrix-ii/discuss/66140/My-concise-O(m%2Bn)-Java-solution 

 

LeetCode All in One 题目讲解汇总(持续更新中...)

[lld2006]

看到一种算法可以将效率提高到log(m)+log(n)的， 核心是二分法

class Solution {
public:

bool searchMatrix(vector<vector<int>>& matrix, int target) {
  if(matrix.empty() || matrix[0].empty()) return false;
    return searchRect(matrix,target,0,0,matrix.size()-1,matrix[0].size()-1);
}

bool searchRect(vector<vector<int>>& matrix, int target, 
                           int top, int left, int bottom, int right) {
    //search if the target is inside the rectangular matrix[top:bottom][left:right]
    //each time we discard 1/4 of all elements
    //time complexity O( log(mn)/log(4/3) ) = O(logm + logn)
    
    if(top>bottom || left>right)
        return false;
    
    int x = (top+bottom)/2;
    int y = (left+right)/2;
    int center = matrix[x][y];
    
    if(center > target){
        return
            searchRect(matrix,target,top,left,x-1,right) ||
            searchRect(matrix,target,x,left,bottom,y-1);
    }
    else if(center < target){
        return
            searchRect(matrix,target,x+1,left,bottom,right) ||
            searchRect(matrix,target,top,y+1,x,right);
    }
    else
        return true;
}

};