#leetcode#Search a 2D Matrix II

标签：leetcode

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.

分析：这题前天刚在EPI上看到....

比 Search a 2D Matrix 多了个条件，那就是每一列都是单调递增的， 假设matrix是 n * n 的， 从右上角元素开始判断， 设为x， 

如果 x == target, 返回 true

如果 x < target, 则向左边的列去寻找

如果 x > target, 则想下面的行去寻找

时间复杂度 O(m + n)

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

#leetcode#Search a 2D Matrix II

标签：leetcode

原文地址：http://blog.csdn.net/chibaoneliuliuni/article/details/47015729