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[leetcode]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 7might become4 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.
二分查找的变形
算法思路：
a[mid] == target return mid;
a[mid] < target 分为两种情况
1. a[mid]和target在同半边，begin = mid + 1
2. a[mid]和target在不同的半边，则a[mid]肯定在后面，target在前半边，因此往前找end = mid - 1;
a[mid] > target同理
1. a[mid]和target在同半边，end = mid - 1;
2. a[mid]和target在不同的半边，则a[mid]肯定在前面，target在后半边，因此往前找begin = mid + 1;
代码如下：
1 public class Solution { 2 public int search(int[] a, int target) { 3 if(a == null || a.length == 0) return -1; 4 int begin = 0, end = a.length - 1; 5 while(begin <= end){ 6 int mid = (begin + end) >> 1; 7 if(a[mid] == target){ 8 return mid; 9 }else if(a[mid] < target){10 if(a[mid] < a[begin] && target >= a[begin]){11 end = mid - 1;12 }else{13 begin = mid + 1;14 }15 }else{16 if(a[mid] >= a[begin] && target < a[begin]){17 begin = mid + 1;18 }else{19 end = mid - 1;20 }21 }22 }23 return -1;24 }25 }
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