原题链接在这里：https://leetcode.com/problems/search-a-2d-matrix-ii/
题目：
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.
题解：
本题是每一行每一列都是升序，从右上角开始x = matrix[i][j], i = 0, j = matrix[0].length - 1, 用x 和target 比较，若是相等则返回true.
若是小于target, 说明肯定不会在这一行，所以i++. 若是大于target, 说明必然不会在这一列所以 j--.
若走到边界还没有返回说明没有这个target, 返回false.
Note: j的初始量是 j = n-1, 注意减一.
Time Complexity: O(m+n). m = matrix.length. n = matrix[0].length.
Space: O(1).
AC Java:
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 i = 0;
int j = n-1; //error
while(i<m && j>=0){
int x = matrix[i][j];
if(x == target){
return true;
}else if(x < target){
i++;
}else{
j--;
}
}
return false;
}
}