Leetcode 74. Search a 2D Matrix

题目链接：Search a 2D Matrix

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 from left to right.
* The first integer of each row is greater than the last integer of the previous row.

　　这道题很简单，为此专门写篇博客其实算博客凑数了。给你一个每一行每一列都是增序，且每一行第一个数都大于上一行末尾数的矩阵，让你判断某个数在这个矩阵中是否存在。
　　假设矩阵是m*n，扫一遍的时间复杂度就是O(m*n)，题目中给出的这么特殊的矩阵，时间复杂度可以降到O(m+n)，具体代码如下，写的比较挫。

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if (matrix.length == 0)
            return false;
        int x = matrix.length;
        int y = matrix[0].length;
        int i = 0, j = 0;
        while (i < x && matrix[i][j] <= target) {
            i++;
        }
        if (i > 0) i--;
        while (j < y) {
            if (j < y && matrix[i][j] == target)
                return true;
            j++;
        }
        return false;
    }
}

文章来源: xindoo.blog.csdn.net，作者：xindoo，版权归原作者所有，如需转载，请联系作者。

原文链接：xindoo.blog.csdn.net/article/details/51869982