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解法一：
利用辅助数组

class Solution {
public:
    /**
     * @param nums: An integer array
     * @return: nothing
     */
    void recoverRotatedSortedArray(vector<int> &nums) {
        // write your code here
        int index = 0;
        int cur = 0;
        for(int i = 1; i < nums.size(); i++){
            if(nums[i] < nums[i-1]){
                index = i;
            }
        }
        int left;
        int right;
        left = 0, right = index - 1;
        while(right >= left){
            int tmp = nums[left];
            nums[left++] = nums[right];
            nums[right--] = tmp;
            
        }
        left = index, right = nums.size() - 1;
        while(right >= left){
            int tmp = nums[left];
            nums[left++] = nums[right];
            nums[right--] = tmp;
        }
        left = 0, right = nums.size() - 1;
        while(right >= left){
            int tmp = nums[left];
            nums[left++] = nums[right];
            nums[right--] = tmp;
        }
    }
};

解法2：利用一个额外的数组即可得出答案。

class Solution {
public:
    /*
     * @param nums: Given an integers array A
     * @return: A long long array B and B[i]= A[0] * ... * A[i-1] * A[i+1] * ... * A[n-1]
     */
    vector<long long> productExcludeItself(vector<int> &nums) {
        // write your code here
        vector<long long> result(nums.size(), 1);
        for(int i = 1; i < nums.size(); i++){
            result[i] = result[i - 1] * nums[i - 1];
        }
        long long temp = 1;
        for(int i = nums.size() - 1; i >= 0; i--){
            result[i] *= temp;
            temp *= nums[i];
        }
        return result;
    }
};