#Leetcode# 33. Search in Rotated Sorted Array

https://leetcode.com/problems/search-in-rotated-sorted-array/

Suppose an array sorted in ascending order 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.

Your algorithm's runtime complexity must be in the order of O(log n).

Example 1:

Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4

Example 2:

Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1

代码：

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int n = nums.size();
        map<int, int> mp;
        mp.clear();
        for(int i = 0; i < n; i ++)
            mp[nums[i]] = i;
        
        sort(nums.begin(), nums.end());
        int l = 0, r = n - 1;
        int mid;
        while(l <= r) {
            mid = (r - l) / 2 + l;
            if(nums[mid] == target) return mp[nums[mid]];
            else if(nums[mid] > target) r = mid - 1;
            else l = mid + 1;
        }
        return -1;
    }
};　

惊险！0.93%

 题解：原来是要找出来旋转的位置 然鹅我只想到记下坐标然后排序再二分 行吧行吧 哭唧唧

 

正解代码：

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int left = 0, right = nums.size() - 1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] == target) return mid;
            else if (nums[mid] < nums[right]) {
                if (nums[mid] < target && nums[right] >= target) left = mid + 1;
                else right = mid - 1;
            } else {
                if (nums[left] <= target && nums[mid] > target) right = mid - 1;
                else left = mid + 1;
            }
        }
        return -1;
    }
};