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 duplic
//时间复杂度O(n) public int search(int[] A, int target) { for(int i=0;i<A.length;i++){ if(target==A[i]) return i; } return -1; }
二分法:
public class Solution { public int search(int[] A,int target){ int first=0,mid,last=A.length; while(first!=last){ mid=(first+last)/2; if(target==A[mid])return mid; if (A[first] <= A[mid]) { if (A[first] <= target && target < A[mid])//有序部分 last = mid; else first = mid + 1; } else { if (A[mid] < target && target <= A[last-1])//有序部分 first = mid + 1; else last = mid; } } return -1; } }