面试经典题---68.文本左右对齐

68.文本左右对齐

我的解法：

两层while循环嵌套，外层循环用于处理不同行，内层while循环计算出每行应有的单词个数。

• 使用left和right记录一行中应有的单词下标，即当前行应包含从words[left]到words[right-1]这count个单词；
• 每行单词之间至少有space/(count-1)个空格，另外每行前space%(count-1)个单词之间应该多一个空格，才能保证空格均匀分配且左侧空格数更多；
• 这里定义了拼接函数join()，用于将words[left]到words[right-1]这些单词以空格space连接起来，注意其中tmp的初值和赋值方式；
• 对于最后一行和只包含一个单词的中间行需要特殊处理

class Solution {
    string join(vector& words, int left, int right, string space){
        string tmp = words[left];
        for(int i = left + 1; i <= right; ++i){
            tmp += space + words[i];
        }
        return tmp;
    }
public:
    vector fullJustify(vector& words, int maxWidth) {
        int n = words.size();
        vector res;
        int right = 0;
        while(true){
            int left = right;
            int len = 0;
            while((right < n) && (words[right].size() + len + right - left <= maxWidth)){
                len += words[right++].size();
            }
            if(right == n){
                string t = join(words, left, n - 1, " ");
                res.push_back(t + string(maxWidth - t.size(), ' '));
                break;
            }
            int count = right - left;
            int space = maxWidth - len;
            if(count == 1){
                res.push_back(words[left] + string(space,' '));
                continue;
            }
            int avg_space = space / (count - 1);
            int extra = space % (count - 1);
            string t1 = join(words, left, left + extra, string(avg_space + 1, ' '));
            string t2 = join(words,left + extra + 1,right - 1, string(avg_space, ' '));
            res.push_back(t1 + string(avg_space,' ') + t2);
        }
        return res;
    }
};