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[leetcode-556-Next Greater Element III]

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[80];
        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 : -1;
    }

 

[leetcode-556-Next Greater Element III]