标签：style blog http color os 2014
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.
题解：还是按照二分的方法找target。
2.如果A[l] >= A[mid], 说明mid以右有序且都大于mid，如下图所示，如果target在mid和r之间，那么需要把l重新置为mid；其他情况都需要到mid左端继续搜索。
当l + 1 = r的时候，只要检查l和r所指向的元素是否等于target即可。
代码如下：
1 public class Solution { 2 public int search(int[] A, int target) { 3 int l = 0; 4 int r = A.length - 1; 5 6 while(l + 1< r){ 7 int mid = l + (r-l)/2; 8 if(A[mid] == target) 9 return mid; 10 if(A[l]< A[mid] ){ 11 if(A[mid] >= target && A[l] <= target) 12 r = mid; 13 else { 14 l = mid; 15 } 16 } 17 else { 18 if(target >= A[mid] && target <= A[r]) 19 l = mid; 20 else { 21 r = mid; 22 } 23 } 24 } 25 26 if(target == A[l]) 27 return l; 28 if(target == A[r]) 29 return r; 30 return -1; 31 } 32 }
【leetcode刷题笔记】Search in Rotated Sorted Array,布布扣,bubuko.com
【leetcode刷题笔记】Search in Rotated Sorted Array
标签：style blog http color os 2014
原文地址：http://www.cnblogs.com/sunshineatnoon/p/3850603.html
