c++ 有swap函数

这是剑指offer数组中重复的数字那个题，直接使用的swap函数

class Solution {
public:
    // Parameters:
    //        numbers:     an array of integers
    //        length:      the length of array numbers
    //        duplication: (Output) the duplicated number in the array number
    // Return value:       true if the input is valid, and there are some duplications in the array number
    //                     otherwise false
    bool duplicate(int numbers[], int length, int* duplication) {
        if(length <= 0){
            *duplication = -1;
            return false;
        }
        int start = 0;
        while(start < length){
            if(numbers[start] == numbers[numbers[start]] && numbers[start] != start){
                *duplication = numbers[start];
                return true;
            }
            if(start == numbers[start]){
                start++;
                continue;
            }
            int index = numbers[start];
            while(numbers[start] != numbers[index]){
                swap(numbers[start],numbers[index]);
                index = numbers[start];
            }
        }
        *duplication = -1;
        return false;
    }
};

字符串的全排列也用到了swap

class Solution {
public:
    vector<string> Permutation(string str) {
        if(str.size() == 0)
            return ans;
        length = str.size();
        int begin = 0;
        Permutation(str,begin);
        //res.clear();
        set<string>::iterator it;
        for (it = res.begin(); it != res.end(); ++it)
            ans.push_back(*it);
        return ans;
    }
    void Permutation(string str,int begin){
        if(begin == length){
            res.insert(str);
            return;
        }
        for(int i = begin;i < length;i++){
            swap(str[begin],str[i]);
            Permutation(str,begin+1);
            swap(str[begin],str[i]);
        }
    }
    set<string> res;
    vector<string> ans;
    int length = 0;
};