Search in Rotated Sorted Array--leetcode

标签：leetcode

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.

class Solution {
public:
    int search(int A[], int n, int target) {
        int left = 0;
        int right = n-1;
        while(left <= right)//left,right均为 mid +/- 1,存在交错，满足退出条件
        {
            int mid = (left+right)/2;
            if(target == A[mid])
                return mid;
            if(A[left] <= A[mid])//mid 在左侧 较大的区间
            {
                if((A[left] <= target)&&(target < A[mid])) //target != A[mid] 上式以排除。target在左侧递增区间
                    right = mid - 1; //target != A[mid],可以让right = mid -1
                else//化为原型式
                    left = mid + 1;// 用mid+1, 存在 left<=mid,  target != A[mid],所以可用left = mid + 1
            }
            else {//mid在右侧区间
                if((A[mid] < target)&&(target <= A[right]))
                    left = mid + 1;
                else//化为原型式
                    right = mid - 1;
            }
        }
        return -1;
    }
};

Search in Rotated Sorted Array--leetcode

标签：leetcode

原文地址：http://blog.csdn.net/youxin2012/article/details/41576717