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].

題意：給一個數字，找出比它小最接近它的 非遞減數 例如 1234，122 ，266等等

這題其實不難，關鍵在找到最近的遞減數的位置，而且我們得倒著去找，為什麽呢？因為會有像是222000這種扯淡的數字，所以我們得一邊找一邊操作

之後就是這個位置之後的數字都改為9就行了

唉，寫的時候被自己蠢哭了，一開始一直都是用String，寫著寫著發現String是不可變的，中間加了一堆中間變量操作

後來想想，用char[] 就可以了啊

class Solution {
    private int toInt(char[] chs) {
        int res = 0;
        for (int i = 0; i < chs.length; i++)
            res = res*10 + (chs[i] - ‘0‘);
        return res;
    }
    public int monotoneIncreasingDigits(int N) {
        String str = String.valueOf(N);
        
 
int n = str.length(), j = n;
        char[] chs = str.toCharArray();
        for (int i = n - 1; i > 0; i--) {
            if (chs[i] < chs[i - 1]) {
                chs[i - 1]--;
                j = i;
            }
        }
        for (int i = j; i < n; i++) {
            chs[i] = ‘9‘;
        }
        return toInt(chs);
    }
}

[LeetCode] 738. Monotone Increasing Digits