556. Next Greater Element III ----------M

 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

A:

就是转成long然后对比呗

class Solution {
public:
    int nextGreaterElement(int n) {
        vector<int> V;
        while(n){
            V.push_back(n%10);
            n = n/10;
        }
        int cDigit = V.size();
        int smallerIndex = -1;
        for(int i =1;i<cDigit;i++){
            if(V[i]<V[i-1]){
                smallerIndex = i;
                break;
            }
        }
        if(smallerIndex<0)
            return -1;
        // first the first place that's smaller than smallerIndex;
        int firstBigger = -1;
        for(int i =0;i<smallerIndex;i++){
            if(V[i]>V[smallerIndex]){
                firstBigger = i;
                break;
            }
        }
        int tmp = V[smallerIndex];
        V[smallerIndex] = V[firstBigger];
        V[firstBigger] = tmp;
        // swap all 
        int start = 0, end = smallerIndex-1;
        while(start<end){
            tmp = V[start];
            V[start] = V[end];
            V[end] = tmp;
            start++;
            end--;
        }
        long res = 0;
        while(not V.empty()){
            int last = V.back();
            V.pop_back();
            res = res * 10 + last;
        }
        if(res > INT_MAX)
            return -1;
        return (int)res;
    }
};