一、题目说明

这个题目是33. Search in Rotated Sorted Array,说的是在一个“扭转”的有序列表中,查找一个元素,时间复杂度O(logn)。

二、我的解答

这是一个查找,根据复杂度,我们知道只能用二分查找。但由于这个不是一个完全的有序列表,故需要改造。先写出二分查找(来源于数据结构):

二分查找:
int search(vector<int>& nums, int target){
if(nums.size()<0) return -1;
int start = 0, end=nums.size()-1;
int mid = -1;
while(start <= end){
//mid = (start + end) /2;
mid = start + (end - start) /2;
if(target == nums[mid]){
return mid;
}else if(target < nums[mid]){
end = mid -1;
}else{
start = mid + 1;
}
}
return -1;
}

然后在二分查找基础上进行改进:

class Solution{
public:
int search(vector<int>& nums, int target){
if(nums.size()<0) return -1;
int start = 0, end=nums.size()-1;
int mid = -1;
while(start<=end){
mid = start + (end - start) / 2;
if(nums[mid] == target) return mid; //前半部分有序
if(nums[start]<=nums[mid]){
//target在前半部分
if(target >=nums[start] && target<nums[mid]){
end = mid -1;
}else{
start = mid + 1;
}
}else{
//后半部分有序
if(target<=nums[end] && target>nums[mid]){
start = mid + 1;
}else{
end = mid -1;
}
}
}
return -1;
}
};

三、优化措施

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