码迷,mamicode.com
首页 > 其他好文 > 详细

leetcode 74. Search a 2D Matrix

时间:2016-08-07 21:23:22      阅读:133      评论:0      收藏:0      [点我收藏+]

标签:

 

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.

 

For example,

Consider the following matrix:

 

[
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]

Given target = 3, return true.

 

分析:

写出一个高效的算法来搜索 m × n矩阵中的值。这个矩阵具有以下特性:

  • 每行中的整数从左到右是排序的。
  • 每行的第一个数大于上一行的最后一个整数。

这道题可以把二维数组里的数字当作是一个有序的列表,用二分法查找。

 

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if (matrix == null || matrix.length == 0) {
            return false;
        }
        if (matrix[0] == null || matrix[0].length == 0) {
            return false;
        }
        
        int row = matrix.length, column = matrix[0].length;
        int start = 0, end = row * column - 1;
        
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            int number = matrix[mid / column][mid % column];
            if (number == target) {
                return true;
            } else if (number < target) {
                start = mid;
            } else {
                end = mid;
            }
        }
        
        if (matrix[start / column][start % column] == target) {
            return true;
        } else if (matrix[end / column][end % column] == target) {
            return true;
        }
        
        return false;
        
    }
}

 

leetcode 74. Search a 2D Matrix

标签:

原文地址:http://www.cnblogs.com/iwangzheng/p/5747278.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!