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：

    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

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。

char * truncateSentence(char * s, int k){
    int n = strlen(s);
    int end = 0, count = 0;
    for (int i = 1; i <= n; i++) {
    //Details:i=1,soi<=n
    //Aim of i=1:To pair with following number of data
    //Meanwhile,because K begins with 1 
        if (i == n || s[i] == ' ') {
    //i==n：It means that the end of string 
    // was choped at the end
    //s[i]==' ':Blank space found
            count++;
            if (count == k) {
    //If's judgment ensures that 
    // the number of choping is in place
    // so break from cycle
                end = i;
                break;
            }
        }
    }
    s[end] = '\0';
    //Enter the terminator after the breaking digit is obtained
    //Since it starts from 1,it can be assigned directly
    return s;
}