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c++ 有swap函数
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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 <= ){
*duplication = -;
return false;
}
int start = ;
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 = -;
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() == )
return ans;
length = str.size();
int begin = ;
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+);
swap(str[begin],str[i]);
}
}
set<string> res;
vector<string> ans;
int length = ;
};
