81. Search in Rotated Sorted Array II

最新推荐文章于 2025-08-08 21:26:37 发布

原创最新推荐文章于 2025-08-08 21:26:37 发布 · 219 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #Search in Rotated Sorted Array

本文介绍了一种在旋转过的有序数组中查找特定元素的算法。该算法通过实现自定义的三等分查找和二分查找方法来高效地解决问题，并针对小规模数组采取了线性查找作为补充。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

和33. Search in Rotated Sorted Array一样。直接上代码

class Solution {

public:
bool search(vector<int>& nums, int target) {

if(nums.size()<1) return false;
if(nums.size()<=6){
for(int i=0;i<nums.size();i++){
if(nums[i]==target){
return true;
}
}
return false;
}
int re =0;
re = thirdSearchPos(nums,0,nums.size() - 1,target);
if(re == -1) return false;
return true;
}

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;
}
};