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leetcode 153: Find Minimum in Rotated Sorted Array

时间:2014-12-30 07:06:33      阅读:166      评论:0      收藏:0      [点我收藏+]

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Find Minimum in Rotated Sorted Array

Total Accepted: 21207 Total Submissions: 65855

Suppose a sorted array is rotated at some pivot unknown to you beforehand.

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

Find the minimum element.

You may assume no duplicate exists in the array.

[分析]

  二分法, 如果 num[mid] 和相邻elements 不是ordered, 最小值可得. 否则, 查看 low 和 high 是否有序, 跳到无顺序的一半.

[注意事项]
NONE

[CODE]

public class Solution {
    public int findMin(int[] num) {
        if(num==null || num.length < 1) return 0; // ask interviewer which value should return: Integer.MIN_VALUE or throw a Exception.
        
        int low = 0, high = num.length-1;
        while(low < high) {
            int mid = low + (high-low)/2;   
            int x = num[mid]; 
            if( mid != num.length-1 && x>num[mid+1]){
                return num[mid+1];                
            } else if(num[mid] < num[low]) {
                high = mid;
            } else if(num[mid] > num[high]) {
                low = mid;
            } else {
                return num[low];
            }
        }
        return num[low];
    }
}


leetcode 153: Find Minimum in Rotated Sorted Array

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原文地址:http://blog.csdn.net/xudli/article/details/42258905

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