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Search in Rotated Sorted Array
Search in Rotated Sorted Array
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.
这道题主要思路：先找到pivot，查找pivot过程中前面的已经查找过，只需对后面的进行二分查找即可
1 public class Solution { 2 public int search(int[] A, int target) { 3 int i = 0; 4 for(; i < A.length; i++){ 5 if(i > 0 && A[i - 1] > A[i]) //找出轴的位置 6 break; 7 if(A[i] == target) 8 return i; 9 }10 if(i == A.length) //没有找到11 return -1;12 //从轴的位置到最后开始二分查找从i的位置开始到length - 113 int low = i;14 int high = A.length - 1;15 int middle;16 while(low <= high){17 middle = (low + high) / 2;18 if(A[middle] == target)19 return middle;20 else if(A[middle] > target){ //前半段找21 high = middle - 1;22 }else{ //后半段找23 low = middle + 1;24 }25 }26 27 return -1;28 }29 }
Search in Rotated Sorted Array
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