leetcode解题: Search a 2D Matrix II (240)

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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:

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[
  [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.

解法1：O(N + M)

这题和Matrix I的区别是上一行和下一行不是递增的，也就是说不能串起来当成一个sorted array来处理。
这题要变换下思路，考虑右上角和左下角两个点。比如左下角，比数字大就往上，比它数字小就往右。直到找到所要求的答案。
Java

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