闯关leetcode——58. Length of Last Word

大纲

题目
- 地址
- 内容
解题
- 代码地址

题目

地址

https://leetcode.com/problems/length-of-last-word/description/

内容

Given a string s consisting of words and spaces, return the length of the last word in the string.

A word is a maximal substring consisting of non-space characters only.

Example 1:

Input: s = “Hello World”
Output: 5
Explanation: The last word is “World” with length 5.

Example 2:

Input: s = " fly me to the moon "
Output: 4
Explanation: The last word is “moon” with length 4.

Example 3:

Input: s = “luffy is still joyboy”
Output: 6
Explanation: The last word is “joyboy” with length 6.

Constraints:

1 <= s.length <= 10⁴
s consists of only English letters and spaces ’ '.
There will be at least one word in s.

解题

这题就是要找到一个字符串中最后一段非空格的连续字符长度。我们只要定义好这样的子串特点即可：以空格后的第一个字符开始，以空格前的最后一个字符结束。向后遍历，找到最后一个符合这个特点的子串。需要注意的是两个边界：我们可以假设原始字符串前有一个空格，原始字符串后有一个空格，这可以保证我们逻辑的一致性。

#include <string>
using namespace std;class Solution {
public:int lengthOfLastWord(string s) {bool preIsSpace = true; // if the previous character is spaceint cur = 0;int lastWordStart = 0;int lastWordEnd = 0;for (char c : s) {if (c == ' ') {if (!preIsSpace) {lastWordEnd = cur;}preIsSpace = true;} else {if (preIsSpace) {lastWordStart = cur;}preIsSpace = false;}cur++;}if (!preIsSpace) {lastWordEnd = cur;}return lastWordEnd - lastWordStart;}
};