leetcode434&58

来源：互联网发布：网络语香菇是什么意思编辑：程序博客网时间：2024/04/30 13:05

1、Number of Segments in a String
Count the number of segments in a string, where a segment is defined to be a contiguous sequence of non-space characters.
Please note that the string does not contain any non-printable characters.
Example:
Input: “Hello, my name is John”
Output: 5

class Solution {public:    int countSegments(string s) {        int res = 0;        for (int i = 0; i < s.size(); i++)             res += s[i] != ' ' && (i + 1 == s.size() || s[i + 1] == ' ');        return res;    }};

计算字符串有几部分组成就是在间断处和最后加一就好了～
间断处就是：s[i] != ’ ‘&&s[i + 1] == ’ ’
最后一处就是：s[i] != ’ ‘&&i + 1 == s.size()

2、Length of Last Word
Given a string s consists of upper/lower-case alphabets and empty space characters ’ ‘, return the length of last word in the string.
If the last word does not exist, return 0.
Note: A word is defined as a character sequence consists of non-space characters only.

For example,
Given s = “Hello World”,
return 5.

class Solution {public:    int lengthOfLastWord(string s) {        if(s.empty()) return 0;        for(auto i=s.end()-1; i>=s.begin(); i--){            if(*i==' ')                s.erase(i);            else                break;        }        for(int i=s.size()-1; i>=0;i--){            if(s[i]==' ')                return s.size()-i-1;        }        return s.size();    }};

跟上题有些类似，判断空格终止循环，但是有几个情况需要额外考虑。
1、最后字符后面存在空格——>从后向前删除最后一个字符后面的所有空格；只有最前面存在空格——>从后向前遍历，找到空格就停啦。
2、判断是否为空，返回0
3、不存在空格，返回字符串长度

0 0