Number of Digit One——LeetCode⑩
//原题链接https://leetcode.com/problems/number-of-digit-one/
- 题目描述
Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n.
Example:
Input: 13 Output: 6 Explanation: Digit 1 occurred in the following numbers: 1, 10, 11, 12, 13.
- 思路分析
1.暴力法:对10求余,判断个位是否为1,然后除10依次判断//会超时
2.优化:总结n每位数的规律,
以n=123为例
(1的个数:count
当前位数上的权重:weight
当前位数上的数字:now
上一位数:last,例如按照从右到左,2的上一位数为3
循环次数:round)
个位数:3属于大于等于1,第十三个循环开始,则count = round + 1 = 12+1;
若为120,即now为小于1,开始第十三个循环,则count = round=12;
十位数及以上位数:以十位数为例,2大于1,则count = round*weight + weight=1*10+10
若为11X,即now为1,则count = round*weight + 1+last=1*10+1+x
若为10X,即now等于0,则count=round*weight=1*10
- 源码附录
class Solution { public int countDigitOne(int n) { if(n<1){ return 0; } int count = 1; for(int i=0;i<=n;i++){ while(i>0){ if(i%10 == 1){ count ++; } i = i / 10; } } return count; } }
class Solution { public int countDigitOne(int n) { if(n<1){ return 0; } int count = 0; int round = n; int weight = 1; int now = 0 ; while(round>0){ now = round%10; round = round/10; if(now==1){ count = count + (n%weight)+1; } else if(now>1){ count = count + weight; } count = count + round*weight; weight = weight*10; } return count; } }