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[Leetcode] Search in Rotated Sorted Array

时间:2014-11-16 07:06:34      阅读:174      评论:0      收藏:0      [点我收藏+]

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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).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

 

Solution:

bubuko.com,布布扣

 1 public class Solution {
 2     public int search(int[] A, int target) {
 3         if(A.length<=0)
 4             return 0;
 5         int start=0;
 6         int end=A.length-1;
 7         while(start<=end){
 8             int mid=(start+end)/2;
 9             if(A[mid]==target)
10                 return mid;
11             if(A[mid]>=A[start]){
12                 if(A[mid]>=target&&target>=A[start]){
13                     end=mid-1;
14                 }
15                 else
16                     start=mid+1;
17             }else if(A[mid]<=A[end]){
18                 if(A[mid]<=target&&target<=A[end]){
19                     start=mid+1;
20                 }else
21                     end=mid-1;
22             }
23         }
24         return -1;
25     }
26 }

Note:

黄色部分,不要少加了=号,这样在某些情况下(比如[1], 0的情况)会进入不了判断条件里,导致start和end无法改变,超时。

[Leetcode] Search in Rotated Sorted Array

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原文地址:http://www.cnblogs.com/Phoebe815/p/4100950.html

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