#leetcode#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

.

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)

leetcode 有一篇这题的讨论，除了这种O（m + n）的解法， 还有更快的..
http://articles.leetcode.com/2010/10/searching-2d-sorted-matrix-part-ii.html

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;
}
}