yankewei / LeetCode

LeetCode 问题的解决方法
MIT License
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1816. 截断句子 #143

Open yankewei opened 2 years ago

yankewei commented 2 years ago

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" 和 "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 前 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:

来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/truncate-sentence 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。

yankewei commented 2 years ago

遍历

没什么说的,很很简单

func truncateSentence(s string, k int) string {
    ret := s
    spaceCount := 0
    for i := 0; i < len(s); i++ {
        if s[i] == ' ' {
            spaceCount++
        }
        if spaceCount == k-1 {
            ret = s[0:i+1]
        }
    }
    return ret
}