[LeetCode] Search in Rotated Sorted Array II [36]

称号

Follow up for "Search in Rotated Sorted Array":
What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

Write a function to determine if a given target is in the array.

解题思路

这题和Search in Rotated Sorted Array问题类似。只是这个题同意元素反复。假设有元素反复，此时採取最保守的策略。一次缩小一个范围。

代码实现

class Solution {
public:
    bool search(int A[], int n, int target) {
        if(A==NULL || n<=0) return false;
        int begin = 0, end = n-1, mid = 0;
        while(begin<=end){
            mid = (begin+end)/2;
            if(A[mid] == target) return true;
            if(A[mid] < A[end]){
                //后半段有序
                if(A[end]>=target && A[mid]<target)
                    begin = mid+1;
                else
                    end =  mid - 1;
            }else if(A[mid] > A[end]){
                //前半段有序
                if(A[begin]<=target && A[mid] > target)
                    end = mid-1;
                else
                    begin = mid+1;
            }else
                // 最保守策略，缩小一个范围
                --end;
        }
        return false;
    }
};