leetcodeOJ 240. Search a 2D Matrix II

Write an efficient algorithm that searches for a value in an m x n 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.

For example,

Consider the following 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]]

Given target = 5, return true.

Given target = 20, return false.

代码如下：

/*思路：每次将矩阵中心的元素与目标值比较，将矩阵分成四部分：左上、左下、右上、右下 */class Solution {public:    bool searchMatrix(vector<vector<int>>& matrix, int target) {        int m = matrix.size();        if(m == 0)            return false;        int n = matrix[0].size();        if(n == 0)            return false;                int up = 0;         int down = m-1;        int left = 0;        int right = n-1;                bool ans = search(matrix, target, up, down, left, right);                return ans;    }private:    bool search(vector<vector<int>>& matrix, int target, int up, int down, int left, int right){        if(up == down && left == right)            return (target == matrix[up][left]);        if(up>down || left > right)            return false;        int midRow = up + (down-up)/2;        int midCol = left + (right-left)/2;        if(target == matrix[midRow][midCol]){            return true;        }        else if(target < matrix[midRow][midCol]){            return search(matrix, target, up, down, left, midCol-1) || search(matrix, target, up, midRow-1, midCol, right);        }        else{            return search(matrix, target, up, midRow, midCol+1, right) || search(matrix, target, midRow+1, down, left, right);        }    }};

另一种比较高效的方法：

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