240. Search a 2D Matrix II -M

Write an efficient algorithm that searches for a value target in an m x n integer matrix matrix. This 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 <= matrix[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

A:

思路是，对角线寻找。  复杂度 O(m+n)

如果从左上角开始，到一个点，如果A[i][i] > target了。 则不可能在右下角的区间。同样。也不能在左上角的区间。

因此只可能在左下角和右上角的区间里

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        int m = matrix.size(), n = matrix[0].size();
        return helper(matrix,target, 0,0,m-1,n-1);
    }
private:
    bool helper(vector<vector<int>>& matrix, int target, int topX, int topY, int botX, int botY){ // bot -> bottom
        if(topX > botX || topY > botY || matrix[topX][topY] > target){
            return false;
        }
        // from [topX, topY] ,going diagnally ,to find a node that's bigger
        int x = topX, y = topY;
        while(x <= botX && y<= botY && matrix[x][y] <= target){
            if(matrix[x][y] == target){
                return true;
            }
            ++x;
            ++y;
        }
        // now matrix[x][y] is bigger than target
        return helper(matrix, target,topX, y, x-1, botY) || helper(matrix, target, x, topY,botX, y-1);
    }
};

当然还有一种方法是逐行查找。 从右上角 开始    复杂度 O(m+n)

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        int m = matrix.size(), n = matrix[0].size();
        int i = 0,j=n-1;
        while(i<m&& j>=0){
            if(matrix[i][j] == target){
                return true;
            }else if(matrix[i][j] < target){
                ++i;
            }else{
                --j;
            }
        }
        return false;
    }
};