Search in Rotated Sorted Array II

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.

思路:解题思路和Search in Rotated Sorted Array差不多,但是出现了另一种情况有重复元素出现,头尾指针各向中间移一步。然后在进行判断。

class Solution {
public:
    bool search(int A[], int n, int target) {
        if(A==NULL||n==0)
            return -1;
        int index1=0;
        int index2=n-1;
        int middle;
        while(index1<index2)
        {
            middle=(index1+index2)/2;
//
if(A[middle]==A[index2]&&A[middle]==A[index1]) { index1++; index2--; } else if(A[middle]>=A[index1]) { if(target>=A[index1]&&target<=A[middle]) index2=middle; else index1=middle+1; } else { if(target<=A[index2]&&target>=A[middle]) index1=middle; else index2=middle-1; } } if(index2>=0&&index2<n&&target==A[index2]) return true; return false; } };
原文地址:https://www.cnblogs.com/awy-blog/p/3637614.html