第9周 Length of Last Word

Length of Last Word

Leetcode algorithms problem 58： Length of Last Word

• 问题描述

  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.

• 问题提示

  A word is defined as a character sequence consists of non-space characters only.

• 思路

  1.一开始拿空字符来判断每个最后单词的长度，采取两个空格之间的长度作为输出，后面发现输入的字符串可以连续空字符，此法作罢
  2.遍历字符串，如果不是空字符，temp变量++，否则变为0，当temp长度不为0时，将值赋给sum

代码

class Solution {public:    int lengthOfLastWord(string s) {        int sum=0;        int temp=0;        for(int i = 0; i < s.length(); i++){            if(s[i] != ' '){                temp++;            }else{                temp = 0;            }            if(temp != 0){                sum = temp;            }        }        return sum;    }};};

时间复杂度: O(n)
空间复杂度: O(1)

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