LeetCode——Search a 2D Matrix II

Description：

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.

首先想到的就是遍历整个矩阵，时间复杂度是O(m*n)，肯定是Timeout。

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        
        for(int i=0; i<matrix.length; i++) {
            for(int j=0; j<matrix[i].length; j++) {
                if(matrix[i][j] == target) {
                    return true;
                }
            }
        }
        
        return false;
    }
}

然后在优化的话就想到了二分。对每一行进行二分。时间复杂度是O(n*logm)，还是Timeout，二分用的越多时间复杂度就越高，所以行列都二分（有点递归分治的意思）更会Timeout。

public class Solution {
    
    public boolean binarySearch(int[] arr, int terget) {
        
        int left = 0, int right = arr.length - 1;
        while(left <= right) {
            int mid = left + (right - left) / 2;
            if(target == arr[mid]) {
                return true;
            }
            if(target > arr[mid]) {
                left = mid + 1;
            }
            else {
                right = mid - 1;
            }
        }
        
        return false;
    }
    
    public boolean searchMatrix(int[][] matrix, int target) {
        
        for(int i=0; i<matrix.length; i++) {
            if(binarySearch(matrix[i], target)) {
                return true;
            }
        }
        
        return false;
    }
    
    
}

这么看来时间复杂度必须在线性的基础上才行。观察一下矩阵不难发现把矩阵逆时针旋转45度类似一棵二叉查找树。所以就可以模仿二叉查找树的方法来做了。

这样的话时间复杂度就是O(m + n)；AC。

public class Solution {
    
    
    public boolean searchMatrix(int[][] matrix, int target) {
        
        if(matrix.length==0 || matrix[0].length==0) {
            return false;
        }
        
        int i = 0, j = matrix[0].length - 1;
        
        while(i < matrix.length && j >= 0) {
            int cur = matrix[i][j];
            if(cur == target) {
                return true;
            }
            else if(cur < target) {
                i ++;
            }
            else {
                j --;
            }
            
        }
        return false;
        
    }
    
}