81. Search in Rotated Sorted Array II

Problem:

Suppose an array sorted in ascending order 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).

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

The array may contain duplicates.

这题是第33题是差不多的，但是多了一个条件，是允许数组内的数是重复的。这样的设定就会让33题的左右界判断出问题，上一题的判断肯定递增的条件是nums[start]<=nums[mid]， 则在[start, mid]上元素是递增的。但是重复会使得=这个条件出现问题，例如[2,3,2,2]这样，所以要把条件划分成<,>和=三种情况。=的时候就把start元素移到下一位，避免判断两个相等的元素。

Code：(LeetCode运行6ms)

class Solution {public:    bool search(vector<int>& nums, int target) {        int start = 0;        int end = nums.size();        while (start != end) {            int mid = (start + end - 1) / 2;            if (nums[mid] == target) {                return true;            }            if (nums[start] < nums[mid]) {                if (nums[start] <= target && target < nums[mid]) {                    end = mid;                } else {                    start = mid + 1;                }            } else if (nums[start] > nums[mid]) {                if (nums[mid] < target && target <= nums[end - 1]) {                    start = mid + 1;                } else {                    end = mid;                }            } else {                start++;            }        }        return false;    }};