Leetcode NO.81 Search in Rotated Sorted Array II

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题目要求如下：

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.

本题题目难度为中等，但是我觉得难度远胜难度为hard的Search in Rotated Sorted Array

本题的代码如下：

class Solution {public:    bool search(int A[], int n, int target) {        int low = 0;        int high = n - 1;        int mid;        while (low <= high) {            mid = (low + high) / 2;            if (A[mid] == target)                return true;            if (A[mid] > A[low]) {                /* left part are sorted */                if (A[mid] > target) {                    if (A[low] > target)                        low = mid + 1;                    else if (A[low] == target)                        return true;                    else                        high = mid - 1;                }                else if (A[mid] < target) {                    low = mid + 1;                }            }            else if (A[mid] < A[low]) {                /* right part are sorted */                if (A[mid] > target) {                    high = mid - 1;                }                else if (A[mid] < target) {                    if (A[high] > target)                        low = mid + 1;                    else if (A[high] == target)                        return true;                    else                        high = mid - 1;                }            }            else                ++low;        }        return false;    }};

这个思路不是我想出来的，是我看讨论区后，觉得这种算法最为简洁：

这个的精髓在与最后一个++low，如果A[mid] == A[low]，证明之间的所有元素都是相等，所以要一只++low，这样处理完之后，就可以判断左边是sorted还是右边是sorted的。。比如11131，如果仅看A[low],A[high],A[mid]，无法判断出哪边是sorted的，其实pivot在右边3后面那个1，所以左边的是完全sorted的，上面描述的处理可以把前面的1都消掉，当然上面那个例子是worst case，时间复杂度也接近O(N)，排除特殊条件后，逻辑如下：

1A[mid] > A[low]:low-mid之间sorted过

1.1 A[mid] > target

1.1.1.A[low] > target: target在mid-high

1.1.2A[low] == target找到

1.1.3A[low] < target: target在low-mid

1.2A[mid] < target:: target在mid-high

2.A[mid] < A[low]:mid-high之间sorted过

2.1 A[mid] > target: target在low-mid

2.2A[mid] < target:

2.2.1 A[high] > target: mid-high之间

2.2.2 A[high] == target:相等

2.2.3 A[high] < target: low-mid之间

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