Search in Rotated Sorted Array II

一、问题描述

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.
二、思路

由于数组中有重复元素，所以我们在循环中需要判断中间元素和末尾元素的大小关系，分三种情况讨论。

三、代码

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