738. Monotone Increasing Digits

Given a non-negative integer N, find the largest number that is less than or equal to N with monotone increasing digits.

(Recall that an integer has monotone increasing digits if and only if each pair of adjacent digits x and y satisfy x <= y.)

Example 1:

Input: N = 10
Output: 9

Example 2:

Input: N = 1234
Output: 1234

Example 3:

Input: N = 332
Output: 299

Note: N is an integer in the range [0, 10^9].

Approach #1: C++. [recursive]

class Solution {
public:
    int monotoneIncreasingDigits(int N) {
        stack<int> temp;

        while (N) {
            int lastNum = N % 10;
            N /= 10;
            temp.push(lastNum);
        }
        
        int curNum = temp.top();
        temp.pop();
        
        int ans = 0;
        while (!temp.empty()) {
            if (curNum <= temp.top()) {
                ans = ans * 10 + curNum;
                curNum = temp.top();
                temp.pop();
            } else {
                ans = ans * 10 + curNum - 1;
                for (int i = 0; i < temp.size(); ++i) 
                    ans = ans * 10 + 9;
                if (judge(ans))
                    return ans;
                else return monotoneIncreasingDigits(ans);
            }
        }
        ans = ans * 10 + curNum;
        return ans;
    }
    
    bool judge(int N) {
        int lastNum = N % 10;
        N /= 10;
        while (N) {
            int curNum = N % 10;
            N /= 10;
            if (curNum > lastNum) return false;
            lastNum = curNum;
        }
        
        return true;
    }
};

　　

Approach #2: Java. [Truncate After Cliff]

class Solution {
    public int monotoneIncreasingDigits(int N) {
        char[] S = String.valueOf(N).toCharArray();
        int i = 1;
        while (i < S.length && S[i-1] <= S[i]) i++;
        while (0 < i && i < S.length && S[i-1] > S[i]) S[--i]--;
        for (int j = i+1; j < S.length; ++j) S[j] = '9';

        return Integer.parseInt(String.valueOf(S));
    }
}