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"

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"

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

空格判断、字符复制

1. 简单，就复制字符
2. 字符复制完了加一个空格
3. 计数，小于k
4. 最后会多复制一个空格最后要截掉

代码

class Solution {
    public String truncateSentence(String s, int k) {
        char[] arr = new char[501];
        int cnt = 0;
        int idx = 0;
        int arrIdx = -1;
        while (cnt < k && idx < s.length()){
            if (s.charAt(idx) == ' '){
                idx++;
                continue;
            }
            while (idx < s.length() && s.charAt(idx) != ' ' ){
                arr[++arrIdx] = s.charAt(idx++);
            }
            arr[++arrIdx] = ' ';
            cnt++;
        }
        char[] arr2 = new char[arrIdx];
        System.arraycopy(arr,0,arr2,0,arrIdx);
        return new String(arr2);
    }
}

其实题目给了限制条件了

 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 

所以可以更简单点，直接判断遇到的空格的个数就可以了

代码

class Solution {
    public String truncateSentence(String s, int k) {
        char[] arr = s.toCharArray();
        int cnt = 0;
        int idx = -1;
        while (cnt < k && ++idx < s.length()){
            if (arr[idx] == ' '){
                cnt++;
            }
        }
        char[] arr2 = new char[idx];
        System.arraycopy(arr,0,arr2,0,idx);
        return new String(arr2);
    }
}