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]
.
题意:给一个数字,找出比它小最接近它的 非递减数 例如 1234,122 ,266等等
这题其实不难,关键在找到最近的递减数的位置,而且我们得倒着去找,为什么呢?因为会有像是222000这种扯淡的数字,所以我们得一边找一边操作
之后就是这个位置之后的数字都改为9就行了
唉,写的时候被自己蠢哭了,一开始一直都是用String,写着写着发现String是不可变的,中间加了一堆中间变量操作
后来想想,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); } }