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.

解题思路： 题目中数组没有重复，则比较A[mid]与A[end]来确定mid所处的位置；然后根据二分查找来修正st和en；确定唯一的情况；

代码：

1. int search(int A[], int n, int target) {
2. if(n<=0) return -1;
3. int st = 0;
4. int en = n-1;
5. while(st<=en) {
6. int mid = st + ((en-st)>>1);
7. if(A[mid] == target) {
8. return mid;
9. }
10. if(A[mid]>A[en]) {
11. if(A[mid] > target && target>=A[st]) { // 不要忘记等号
12. en = mid - 1;
13. } else {
14. st = mid + 1;
15. }
16. } else {
17. if(A[mid] < target && target <= A[en]) { //不要忘记等号
18. st = mid + 1;
19. } else {
20. en = mid - 1;
21. }
22. }
23. }
24. return -1;
25. }

递归实现：

1. int search(int A[], int n, int target) {
2. return search(A,0,n-1,target);
3. }
4. int search(int A[],int start,int end,int target) {
5. if(start>end) return -1;
6. int mid = start + ((end - start) >> 1); // + has high priority than >>
7. int num1 = A[start];
8. int num2 = A[mid];
9. int num3 = A[end];
10. if(num2 == target) return mid;
11. if(num1 <= num2) {
12. if(num1 > target || target > num2) {
13. return search(A,mid+1,end,target);
14. } else {
15. return search(A,start,mid-1,target);
16. }
17. }
18. if(target < num2 || target > num3)
19. return search(A,start,mid-1,target);
20. return search(A,mid+1,end,target);
21. }