leetcode:Compare Version Numbers
2015-01-15 08:57
381 查看
题目描述:
Compare two version numbers version1 and version1.
If version1 > version2 return 1, if version1 < version2 return -1, otherwise return 0.
You may assume that the version strings are non-empty and contain only digits and the
The
For instance,
second-level revision of the second first-level revision.
Here is an example of version numbers ordering:
Credits:
Special thanks to @ts for adding this problem and creating all test cases.
这个首先想到了android种版本升级的那个,可惜andorid用的是versioncode 而不是versionname是很简单的整数判断,不过也是很实际的例子。
考虑的情况首先是以分隔符“.”为例的 1中 1.1 2.1这种从前往后开始遍历时 分隔符之前就能判断出1.1是小于2.1的 return -1;
对于 1.1 和 1.2 当分隔符之前相等 此时重新判断分隔符之后内容 1.1<1.2 return -1 ;
对于更多影响的元素 譬如 0.2.1 1.1或者 1.3.1 1.3 ,1.2.0.1 1.1.2.3,等即保持从前往后遍历,遇到分隔符“.”即开始分割比较,是可以比较出结果的。
accepted code with c++:
Compare two version numbers version1 and version1.
If version1 > version2 return 1, if version1 < version2 return -1, otherwise return 0.
You may assume that the version strings are non-empty and contain only digits and the
.character.
The
.character does not represent a decimal point and is used to separate number sequences.
For instance,
2.5is not "two and a half" or "half way to version three", it is the fifth
second-level revision of the second first-level revision.
Here is an example of version numbers ordering:
0.1 < 1.1 < 1.2 < 13.37
Credits:
Special thanks to @ts for adding this problem and creating all test cases.
这个首先想到了android种版本升级的那个,可惜andorid用的是versioncode 而不是versionname是很简单的整数判断,不过也是很实际的例子。
考虑的情况首先是以分隔符“.”为例的 1中 1.1 2.1这种从前往后开始遍历时 分隔符之前就能判断出1.1是小于2.1的 return -1;
对于 1.1 和 1.2 当分隔符之前相等 此时重新判断分隔符之后内容 1.1<1.2 return -1 ;
对于更多影响的元素 譬如 0.2.1 1.1或者 1.3.1 1.3 ,1.2.0.1 1.1.2.3,等即保持从前往后遍历,遇到分隔符“.”即开始分割比较,是可以比较出结果的。
accepted code with c++:
class Solution { public: int compareVersion(string version1, string version2) { int temp1 = 0; int temp2 = 0; string::iterator t1 = version1.begin(); string::iterator t2 = version2.begin(); while(t1 != version1.end() || t2 != version2.end()) { while(*t1 != '.' && t1 != version1.end()) { temp1 = temp1 *10 + *t1++ - '0'; } while(*t2 != '.' && t2 != version2.end()) { temp2 = temp2 *10 + *t2++ - '0'; } if(temp1 > temp2) return 1; else if(temp1 < temp2) return -1; else { if(t1 == version1.end() && t2 == version2.end()) return 0; else if(t1 == version1.end() && t2 != version2.end()) { temp1 = 0; //这里用到的一个技巧即将分隔符之前已经比较出来的结果重置,无需再对temp1,temp2相加,影响后面的运算 temp2 = 0; t2++; } else if(t1 != version1.end() && t2 == version2.end()) { temp1 = 0; temp2 = 0; t1++; } else { temp1 = 0; temp2 = 0; t1++; t2++; } } } return 0; } };
相关文章推荐
- [LeetCode] Compare Version Numbers
- LeetCode: Compare Version Numbers
- 【Leetcode】【Easy】Compare Version Numbers
- Leetcode: Compare Version Numbers
- LeetCode 165 Compare Version Numbers
- LeetCode Compare Version Numbers, Maximum Gap, Find Peak Element, Intersection of Two Linked Lists
- [Leetcode]Compare Version Numbers
- leetcode由易入难——【5】Compare Version Numbers
- [LeetCode] Compare Version Numbers
- LeetCode题解——Compare Version Numbers
- LeetCode Compare Version Numbers
- leetcode Compare Version Numbers
- LeetCode(68)-Compare Version Numbers
- [LeetCode] Compare Version Numbers
- Leetcode: Compare Version Numbers
- [LeetCode]Compare Version Numbers
- 【Leetcode】Compare Version Numbers
- LeetCode(68)-Compare Version Numbers
- leetcode[165] Compare Version Numbers
- LeetCode[String]: Compare Version Numbers