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English Version

题目描述

句子 是一个单词列表,列表中的单词之间用单个空格隔开,且不存在前导或尾随空格。每个单词仅由大小写英文字母组成(不含标点符号)。

  • 例如,"Hello World""HELLO""hello world hello world" 都是句子。

给你一个句子 s​​​​​​ 和一个整数 k​​​​​​ ,请你将 s​​ 截断 ​,​​​使截断后的句子仅含 k​​​​​​ 个单词。返回 截断 s​​​​​​ 后得到的句子

 

示例 1:

输入:s = "Hello how are you Contestant", k = 4
输出:"Hello how are you"
解释:
s 中的单词为 ["Hello", "how" "are", "you", "Contestant"]
前 4 个单词为 ["Hello", "how", "are", "you"]
因此,应当返回 "Hello how are you"

示例 2:

输入:s = "What is the solution to this problem", k = 4
输出:"What is the solution"
解释:
s 中的单词为 ["What", "is" "the", "solution", "to", "this", "problem"]
前 4 个单词为 ["What", "is", "the", "solution"]
因此,应当返回 "What is the solution"

示例 3:

输入:s = "chopper is not a tanuki", k = 5
输出:"chopper is not a tanuki"

 

提示:

  • 1 <= s.length <= 500
  • k 的取值范围是 [1,  s 中单词的数目]
  • s 仅由大小写英文字母和空格组成
  • s 中的单词之间由单个空格隔开
  • 不存在前导或尾随空格

解法

Python3

class Solution:
    def truncateSentence(self, s: str, k: int) -> str:
        return ' '.join(s.split()[:k])
class Solution:
    def truncateSentence(self, s: str, k: int) -> str:
        for i, c in enumerate(s):
            if c == ' ':
                k -= 1
            if k == 0:
                return s[:i]
        return s

Java

class Solution {
    public String truncateSentence(String s, int k) {
        for (int i = 0; i < s.length(); ++i) {
            if (s.charAt(i) == ' ' && (--k) == 0) {
                return s.substring(0, i);
            }
        }
        return s;
    }
}

C++

class Solution {
public:
    string truncateSentence(string s, int k) {
        for (int i = 0; i < s.size(); ++i) {
            if (s[i] == ' ' && (--k) == 0) {
                return s.substr(0, i);
            }
        }
        return s;
    }
};

Go

func truncateSentence(s string, k int) string {
	for i, c := range s {
		if c == ' ' {
			k--
		}
		if k == 0 {
			return s[:i]
		}
	}
	return s
}

JavaScript

/**
 * @param {string} s
 * @param {number} k
 * @return {string}
 */
var truncateSentence = function (s, k) {
    for (let i = 0; i < s.length; ++i) {
        if (s[i] == ' ' && --k == 0) {
            return s.slice(0, i);
        }
    }
    return s;
};

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