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我的解法#xff1a; 两层while循环嵌套#xff0c;外层循环用于处理不同行#xff0c;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#xff0c;即当前行应包含从words[left]到words[right-1]这count个单词#xf…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(vectorstring 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:vectorstring fullJustify(vectorstring words, int maxWidth) {int n words.size();vectorstring 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;}
};
