版本比较、One Edit Distance 一个编辑距离、Reverse Words in a String 翻转字符串中的单词、single number
1.
Compare Version Numbers 版本比较
Compare two version numbers version1 and version2.
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, 2.5 is not "two and a half" or "half way to version three", it is the fifth second-level revision of the second first-level revision.
Example 1:
Input:version1= "0.1",version2= "1.1" Output: -1
Example 2:
Input:version1= "1.0.1",version2= "1" Output: 1
Example 3:
Input:version1= "7.5.2.4",version2= "7.5.3" Output: -1
class Solution {
public:
int compareVersion(string version1, string version2) {
int n1 = version1.size(), n2 = version2.size();
int i = 0, j = 0, d1 = 0, d2 = 0;
string v1, v2;
while (i < n1 || j < n2) {
while (i < n1 && version1[i] != '.') {
v1.push_back(version1[i++]);
}
d1 = atoi(v1.c_str());
while (j < n2 && version2[j] != '.') {
v2.push_back(version2[j++]);
}
d2 = atoi(v2.c_str());
if (d1 > d2) return 1;
else if (d1 < d2) return -1;
v1.clear(); v2.clear();
++i; ++j;
}
return 0;
}
};
2.One Edit Distance 一个编辑距离
Given two strings S and T, determine if they are both one edit distance apart.
这道题是之前那道Edit Distance的拓展，然而这道题并没有那道题难，这道题只让我们判断两个字符串的编辑距离是否为1，那么我们只需分下列三种情况来考虑就行了：
1. 两个字符串的长度之差大于1，那么直接返回False
2. 两个字符串的长度之差等于1，那么长的那个字符串去掉一个字符，剩下的应该和短的字符串相同
3. 两个字符串的长度之差等于0，那么两个字符串对应位置的字符只能有一处不同。
分析清楚了所有的情况，代码就很好写了，参见如下：
class Solution {
public:
bool isOneEditDistance(string s, string t) {
if (s.size() < t.size()) swap(s, t);
int m = s.size(), n = t.size(), diff = m - n;
if (diff >= 2) return false;
else if (diff == 1) {
for (int i = 0; i < n; ++i) {
if (s[i] != t[i]) {
return s.substr(i + 1) == t.substr(i);
}
}
return true;
} else {
int cnt = 0;
for (int i = 0; i < m; ++i) {
if (s[i] != t[i]) ++cnt;
}
return cnt == 1;
}
}
};3. Reverse Words in a String 翻转字符串中的单词
Given an input string, reverse the string word by word.
For example,
Given s = "the sky is blue",
return "blue is sky the".
Update (2015-02-12):
For C programmers: Try to solve it in-place in O(1) space.
click to show clarification.
Clarification:
- What constitutes a word?
A sequence of non-space characters constitutes a word. - Could the input string contain leading or trailing spaces?
Yes. However, your reversed string should not contain leading or trailing spaces. - How about multiple spaces between two words?
Reduce them to a single space in the reversed string.
用字符串流类stringstream的解法，我们先把字符串装载入字符串流中，然后定义一个临时变量tmp，然后把第一个单词赋给s，这里需要注意的是，如果含有非空格字符，那么每次>>操作就会提取连在一起的非空格字符，那么我们每次将其加在s前面即可；如果原字符串为空，那么就不会进入while循环；如果原字符串为许多空格字符连在一起，那么第一个>>操作就会提取出这些空格字符放入s中，然后不进入while循环，这时候我们只要判断一下s的首字符是否为空格字符，是的话就将s清空即可，参见代码如下：
class Solution {
public:
void reverseWords(string &s) {
istringstream is(s);
string tmp;
is >> s;
while(is >> tmp) s = tmp + " " + s;
if(!s.empty() && s[0] == ' ') s = "";
}
};4.Single Number II 单独的数字之二
Given an array of integers, every element appears three times except for one. Find that single one.
Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
这道题是之前那道 Single Number 单独的数字 的延伸，
利用计算机按位储存数字的特性来做的，这道题就是除了一个单独的数字之外，数组中其他的数字都出现了三次，那么还是要利用位操作 Bit Operation 来解此题。我们可以建立一个32位的数字，来统计每一位上1出现的个数，我们知道如果某一位上为1的话，那么如果该整数出现了三次，对3去余为0，我们把每个数的对应位都加起来对3取余，最终剩下来的那个数就是单独的数字。代码如下：
class Solution {
public:
int singleNumber(vector& nums) {
int res = 0;
for (int i = 0; i < 32; ++i) {
int sum = 0;
for (int j = 0; j < nums.size(); ++j) {
sum += (nums[j] >> i) & 1;
}
res |= (sum % 3) << i;
}
return res;
}
};