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.

本题目，开始的时候想法是遍历每一行，每一行使用binary search来做，时间复杂度是O(mlongn)。代码如下：

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if(matrix==null||matrix.length==0||matrix[0].length==0) return false;
        int m = matrix.length;
        int n = matrix[0].length;
        for(int i=0;i<m;i++){
            int left = 0,right = n-1;
            while(left<=right){
                int mid = left+(right-left)/2;
                if(matrix[i][mid]==target) return true;
                else if(matrix[i][mid]<target) left = mid+1;
                else right = mid-1;
            }
        }
        return false;
    }
}
//the run time complexity could be O(mlongn), the space complexity could be O(1);

看了答案才知道原来可以达到O(m+n)来做，代码如下：

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if(matrix==null||matrix.length==0||matrix[0].length==0) return false;
        int m = matrix.length;
        int n = matrix[0].length;
        int row = 0;
        int col = n-1;
        while(row<m&&col>=0){
            if(matrix[row][col]==target) return true;
            else if(matrix[row][col]<target&&row<m-1){
                row++;
            }else if(matrix[row][col]>target&&col>0){
                col--;
            }else{
                break;
            }
        }
        return false;
    }
}
//the run time complexity would be O(m+n),the space complexity could be O(1);