【LeetCode】81. Search in Rotated Sorted Array II (2 solutions)

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.

与Find Minimum in Rotated Sorted Array,Find Minimum in Rotated Sorted Array II,Search in Rotated Sorted Array对照看

解法一：顺序查找

class Solution {
public:
    bool search(int A[], int n, int target) {
        for(int i = ; i < n; i ++)
        {
            if(A[i] == target)
                return true;
        }
        return false;
    }
};

解法二：二分查找

关键点在于，如果mid元素与low或者high元素相同，则删除一个low或者high

class Solution {

public:
    bool search(int A[], int n, int target) {
        int low = ;
        int high = n-;
        while (low <= high)
        {
            int mid = (low+high)/;
            if(A[mid] == target)
                return true;
            if (A[low] < A[mid])
            {
                if(A[low] <= target && target < A[mid])
                //binary search in sorted A[low~mid-1]
                    high = mid - ;
                else
                //subproblem from low to high
                    low = mid + ;
            }
            else if(A[mid] < A[high])
            {
                if(A[mid] < target && target <= A[high])
                //binary search in sorted A[mid+1~high]
                    low = mid + ;
                else
                //subproblem from low to mid-1
                    high = mid - ;
            }
            else if(A[low] == A[mid])
                low += ;    //A[low]==A[mid] is not the target, so remove it
            else if(A[mid] == A[high])
                high -= ;  //A[high]==A[mid] is not the target, so remove it
        }
        return false;
    }
};