Search in Rotated Sorted Array
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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.
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思路：此题算法上不难，第一步计算旋转的步长。然后根据步长分情况得到升序的新的数组。然后二分查找target的索引，找到后再分情况返回元素的位置。
具体代码如下：
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public class Solution {
public int search(int[] nums, int target) {
if(nums.length == 0){
return -1;
}
if(nums.length == 1){
if(nums[0] == target)
return 0;
return -1;
}
int rotate = 0;//旋转的步长
for(int i = nums.length - 1; i > 0; i--){
if(nums[i] < nums[i-1]){
rotate = i;
break;
}
}
int[] a = new int[nums.length];
//将数组按升序填充到新数组
for(int i = rotate; i < nums.length; i++){
a[i - rotate] = nums[i];//后面未旋转部分
}
for(int i = 0; i < rotate; i++){
a[i+nums.length-rotate] = nums[i];//前面旋转部分
}
int index = -1;
//二分查找
int low = 0;
int hight = nums.length - 1;
while(low <= hight){
int mid = (low + hight)/2;
if(a[mid] > target){
hight = mid - 1;
}else if(a[mid] == target){
index = mid;
break;
}else{
low = mid + 1;
}
}
if(index == -1)
return -1;
if(index + rotate > nums.length - 1)
return index + rotate - nums.length;
return index + rotate;
}
}
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