【LeetCode】33. Search in Rotated Sorted Array 解题小结

题目：

Suppose a sorted array is rotated at some pivot unknown to you beforehand.

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

读入一个元素就将这个元素插入unordered_map里，然后判断是否在里面。

class Solution {
public:int search(vector<int>& nums, int target) {unordered_map<int, int> map;for(int i = 0; i < nums.size(); i++){map[nums[i]] = i;if(map.find(target)!=map.end()){return map[nums[i]];}}return -1;}
};

后面看看提交情况，发现自己还是too young too simple，本题考查的还是二分法搜索。

class Solution
{
public:int search(vector<int>& nums, int target){int left = 0, right = nums.size()-1;while(left <= right){int middle = left + ((right-left)>>1);if (nums[middle] == target) return middle;if (nums[middle] < nums[right]){if (nums[middle] < target && target <= nums[right])left = middle + 1;else  right = middle - 1;}else if (nums[middle] > nums[right]){if (nums[left] <= target && target < nums[middle])right = middle - 1;else left = middle + 1;}else  right--;}return -1;}
};

转载于:https://www.cnblogs.com/Doctengineer/p/5812536.html

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