求出[1,2,3,…,n]全排列的第k个数
纯暴力为超时,其实每隔一个数都可以排除一些数,不用每个数每个数的向上搜索。
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class Solution {
public:
string getPermutation(int n, int k) {
count = 0;
flag = new bool[n+1];
for(int i = 1;i <= n;i++) {
flag[i] = false;
}
string ans = "";
string result = "";
for(int i = 1;i <= n;i++) {
if(result != "") break;
if(k > factorial(n-1) + count) {
count += factorial(n-1);
continue;
}
if(flag[i] == false) {
flag[i] = true;
ans.push_back(i + '0');
getPermutationStep(n, k, 1, ans);
ans.erase(ans.end()-1);
flag[i] = false;
}
}
return result;
}
private:
int count;
bool *flag;
string result;
void getPermutationStep(int n, int k, int now, string ans) {
if(result != "") return;
if(now == n) {
count++;
if(k <= count) {
result = ans;
}
return;
}
for(int i = 1;i <= n;i++) {
if(flag[i] == false) {
if(now==1 && k > factorial(n-now-1) + count) {
count += factorial(n-now-1);
continue;
}
flag[i] = true;
ans.push_back(i + '0');
getPermutationStep(n, k, now+1, ans);
ans.erase(ans.end()-1);
flag[i] = false;
}
}
}
}
int factorial(int x) {
int ans = 1;
for(int i = 1;i <= x;i++) {
ans *= i;
}
return ans;
}
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