leetcode个人题解——#33 Search in Rotated Sorted Array

思路：每次取中间元素，一定有一半有序，另一半部分有序，有序的部分进行二分查找，部分有序的部分递归继续处理。

class Solution {
public:
    int pos = -;
    int middleSearch(int left, int right,vector<int>& nums, int target){//二分查找
        if(left > right || right > nums.size() -  || left < )return -;
        if(nums[left] == target)return left;
        int mid = (left + right) / ;
        if(nums[mid] == target)return mid;
        if(target>nums[mid]) return middleSearch(mid+,right,nums,target);
        else
            return middleSearch(left,mid - ,nums,target);

    }

    void mixSearch(int left, int right, vector<int>& nums, int target){//混合处理
        if(left > right || right > nums.size() -  || left < ) return;
        if(nums[left] == target) {
            pos = left;
            return;
        }
        int ans;
        int middle = (left + right) / ;
        if(nums[middle] > nums[right]) {
            ans = middleSearch(left,middle,nums,target);
            if(ans == -) mixSearch(middle+,right,nums,target);
            else{
                pos = ans;
                return;
            }
        }
        else{
            ans = middleSearch(middle,right,nums,target);
            if(ans == -) mixSearch(left,middle - ,nums,target);
            else{
                pos = ans;
                return;
            }
        }
    }

    int search(vector<int>& nums, int target) {
        mixSearch(,nums.size() - ,nums,target);
        return pos;
    }
};

写的冗余了点，附上简洁的代码：

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int l = , r = nums.size() - ;

        while(l <= r){
            int mid = l + (r - l) / ;

            if(nums[mid] == target) return mid;
            if(nums[mid] > nums[r]){
                if(target > nums[mid] || target <= nums[r]) l = mid + ;   // condition for pick right side
                else r = mid - ;    // else, pick left side
            }else{
                if(target <= nums[r] && target > nums[mid]) l = mid + ;  // condition for pick right side
                else r = mid - ;     // else, pick left side
            }
        }

        return -;
    }
};