leetcode 33. Search in Rotated Sorted Array

最新推荐文章于 2024-10-23 14:58:41 发布

站在风口的熊

最新推荐文章于 2024-10-23 14:58:41 发布

阅读量152

点赞数

CC 4.0 BY-SA版权

分类专栏： leetcode&poj 文章标签： leetcode

本文链接：https://blog.youkuaiyun.com/barebear007/article/details/78523772

leetcode&poj 专栏收录该内容

83 篇文章

订阅专栏

本文详细解析了LeetCode第33题“在旋转排序数组中搜索”的解决方案，通过三等分查找和二分查找相结合的方法，在复杂度较高的情况下仍能有效地找到目标元素。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

leetcode 33. Search in Rotated Sorted Array

class Solution {
public:
    int search(vector<int>& nums, int target) {
        if(nums.size()<1) return -1;
        if(nums.size()<=6){
            for(int i=0;i<nums.size();i++){
                if(nums[i]==target){
                    return i;
                }
            }
        }
        int re =0;
        re = thirdSearchPos(nums,0,nums.size() - 1,target);
        return re;
    }
    int thirdSearchPos(vector<int>& nums,int left, int right,int target) {
        if(right - left<=6){
            for(int i=0;i<nums.size();i++){
                if(nums[i]==target){
                    return i;
                }
            }
            return -1;
        }
        int firstThird = left + (right - left)/3;
        int secondThird = firstThird + (right - left)/3;
        if(target==nums[left]) return left;
        if(target==nums[firstThird]) return firstThird;
        if(target==nums[secondThird]) return secondThird;
        if(target==nums[right]) return right;
        int re = -1;
        if(nums[left]<=nums[firstThird] && target>nums[left] && target<nums[firstThird]){
             re = binarySearchPos(nums,left,firstThird,target);
        }else if(target>nums[left] && target<nums[firstThird]){

            re = thirdSearchPos(nums,left,firstThird,target);
        } 
        if(re!=-1) return re;

        if(nums[secondThird]<=nums[right] && target>nums[secondThird] && target<nums[right]){
            re =  binarySearchPos(nums,secondThird,right,target);
        }else if(target>nums[secondThird] && target<nums[right]){

            re =  thirdSearchPos(nums,secondThird,right, target);
        }   

        if(re!=-1) return re;
        return thirdSearchPos(nums,firstThird,secondThird, target);

    }

      int binarySearchPos(vector<int>& nums1, int start, int end, int num) {
        int left = start, right = end, medium = 0;
        while (left<right) {
            medium = left + (right - left) / 2;
            if (nums1[medium] - num >0) {
                right = medium;
            }
            else if (num - nums1[medium] >0) {

                left = medium + 1;
            }
            else {
                return medium;
            }


        }
        int left1 = -1;
        if (nums1[left] == num)
            left1 = left;

        return  left1;
    }
};