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 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;
	}
}

  

原文地址:https://www.cnblogs.com/gaoxiangde/p/4357275.html