[LeetCode]Search in Rotated Sorted Array II

题目描述:(链接)

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.

解题思路 ：

需要跳过重复的元素。

class Solution {
public:
    bool search(vector<int>& nums, int target) {
        size_t first = 0;
        size_t last = nums.size();
        
        while (first != last) {
            size_t mid = first + (last - first) / 2;
            if (nums[mid] == target) {
                return true;
            } else if (nums[mid] > nums[first]) { // 递增
                if (nums[first] <= target && target < nums[mid]) {
                    last = mid;
                } else {
                    first = mid + 1;
                }
            } else if (nums[mid] < nums[first]) {//递减
                if (nums[mid] < target && target <= nums[last - 1]) {
                    first = mid + 1;
                } else {
                    last = mid;
                }
            } else {// 跳过重复元素
                ++first;
            }
        }
        
        return false;
    }
};