LeetCode 刷题【68. 文本左右对齐】

发布于:2025-09-08 ⋅ 阅读:(29) ⋅ 点赞:(0)

68. 文本左右对齐

自己做

解:直接遍历划分情况

class Solution {
public:
    vector<string> fullJustify(vector<string>& words, int maxWidth) {
        vector<string> res;
        string row;                           //当前行
        int row_len = 0;                      //当前行的单词长度

        int begin = 0;                        //begin,end标记一行起始
        int end = 0;

        while(end < (int)words.size()){
            row_len += words[end].size() + 1;

            if(row_len - 1 > maxWidth){               //超过一行限制,该行包含的单词:[begin,end) => 从words[begin]取到words[end - 1],注:这里必然不可能是尾行
                row_len -= (words[end].size() + 2);        //去除越界的部分:最后一个单词和两边的空格
                // cout << row_len << endl;
                int space_sum = maxWidth - row_len;        //要补充的空格总数
                int div_space = 0;
                int add_space = 0;

                if(end - begin - 1 != 0){
                    div_space = space_sum / (end - begin - 1);       //每个间隔的补充长度
                    add_space = space_sum % (end - begin - 1);       //额外补充间隔的长度             
                }

                // cout << begin << "," << end << "," << div_space << "," << add_space << endl;

                //一行
                for(int i = begin; i < end; i++){
                    row += words[i];                      //拼接单词
                    
                    //加空格
                    if(i != end - 1){
                        row += string(div_space + 1, ' ');   //每个间隔长度

                        if(add_space > 0){                   //如果有额外要补充的间隔,从左往右进行补充
                            row += ' ';
                            add_space--;
                        }
                    }
                }

                //一行只有一个单词的情况,往后补充空格
                if(begin == end - 1)
                    //尾部加空格
                    row += string(maxWidth - row_len, ' ');
                    
                //一行结束,调整begin起始位置,将该行加入结果
                res.push_back(row);
                begin = end;
                row = "";
                row_len = 0;

            }
            else if(row_len - 1 == maxWidth){     //刚好够一行,该行包含的单词:[begin,end] => 从words[begin]取到words[end]
                //一行
                for(int i = begin; i <= end; i++){
                    row += words[i];                      //拼接单词
                    
                    //加空格
                    if(i != end)
                        row += " ";                           //每个间隔长度为1
                }

                //一行结束,调整begin起始位置,将该行加入结果
                res.push_back(row);
                end++;
                begin = end;
                row = "";
                row_len = 0;

            }
            else{                           //对于没填充满的情况: row_len - 1 < 16
                if(end == (int)words.size() - 1){          //如果是尾行,后面补充空格
                    //一行
                    for(int i = begin; i <= end; i++){
                        row += words[i];                      //拼接单词
                        
                        //加空格
                        row += " ";                           //每个间隔长度为1
                    }

                    //尾部加空格
                    row += string(maxWidth - row_len, ' ');

                    //加入结果并结束循环
                    res.push_back(row);
                    break;                                      
                }
                else                                        //如果不是尾行,继续往后延伸
                    end++;

            }
        }

        return res;
    }
};


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