Given a positive 32-bit integer n, you need to find the smallest 32-bit integer which has exactly the same digits existing in the integer n and is greater in value than n. If no such positive 32-bit integer exists, you need to return -1.
Example 1:
Input: 12
Output: 21
Example 2:
Input: 21
Output: -1
思路:
首先,将整型数字转换成字符串,然后利用stl提供的next_permutation()函数,求字符的全排列,对应的字符串再转换回整型,随时记录大小即可。
int nextGreaterElement2(int n) {
char buf[];
sprintf(buf, "%d", n);
string s = buf;
puts(s.data());
sort(s.begin(), s.end());
long long ans = INT_MAX + 1LL;
do {
long long tmp = atoll(s.c_str());
if (tmp > n) {
ans = min(ans, tmp);
}
} while (next_permutation(s.begin(), s.end()));
return ans <= INT_MAX ? ans : -;
}