[Leetcode]Reverse Integer
2014-10-28 07:16
253 查看
核心思想:原数对10取余数赋值给新数后降一位,再把新数升一位加上下一次原数取余值,直到原数降为0。
解法如下:
解法如下:
int reverse(int x) { bool minus = false; if(x<0) { x= -x; minus = true; } int a=0; while(x) { a*=10; a+=x%10; x/=10; } return minus?-a:a;
相关文章推荐
- leetcode 7 Reverse Integer
- leetcode 7 Reverse Integer
- 【LeetCode】7. Reverse Integer
- leetcode 104: Reverse Integer
- [LeetCode]Reverse Integer
- [LeetCode]Reverse Integer
- leetcode--Reverse Integer
- leetcode_7_Reverse Integer
- Leetcode(18) - Reverse Integer
- LeetCode7 Reverse Integer
- [leetcode][javascript]Reverse Integer
- LeetCode : Reverse Integer
- Leetcode_num4_Reverse Integer
- [leetcode] Reverse Integer
- Leetcode学习(24)—— Reverse Integer
- LeetCode 7 - Reverse Integer
- [leetcode]Reverse Integer
- LeetCode_Reverse Integer
- LeetCode--Reverse Integer
- leetcode--Reverse integer