假设我们有一个单词列表和一个宽度k,我们必须对文本进行排列,以使每行恰好有k个字符,并且文本是完全合理的。在这里,我们将打包我们的单词,并在每一行中插入尽可能多的单词。并且我们将在必要时填充多余的空格'',以便每行恰好有k个字符。
在此,单词之间的多余空间应尽可能均匀地分布。如果一行中的空格数量在单词之间分配不均,则左侧的空插槽将比右侧的插槽分配更多的空间。对于文本的最后一行,应保留其两端对齐,并且单词之间不应插入多余的空格。
因此,如果输入类似于[“ The”,“ grumpy”,“ wizards”,“ make”,“ toxic”,“ brew”,“ for”,“ the”,“ evil”,“ queen”,“ and” ,“ Jack”]和k = 13
那么输出将是-
The grumpy wizards make toxic brew for the evil queen and Jack
让我们看下面的实现以更好地理解-
#include <bits/stdc++.h>
using namespace std;
void print_vector(vector<auto> v){
cout << "[";
for(int i = 0; i<v.size(); i++){
cout << v[i] << ", ";
}
cout << "]"<<endl;
}
void print_vector(vector<vector<auto> > v){
cout << "[";
for(int i = 0; i<v.size(); i++){
cout << "[";
for(int j = 0; j <v[i].size(); j++){
cout << v[i][j] << ", ";
}
cout << "],";
}
cout << "]"<<endl;
}
class Solution {
public:
vector<string> fullJustify(vector<string> &a, int b) {
vector <string> result;
int i, j;
for(i = 0; i < a.size(); i = j){
int width = 0;
for(j = i; j < a.size() && width + a[j].size() + j - i <= b; j++){
width += a[j].size();
}
int space = 1;
int extra = 0;
if(j - i != 1 && j != a.size()){
space = (b - width) / (j - i - 1);
extra = (b - width) % (j - i - 1);
}
string line(a[i]);
for(int k = i + 1; k < j; k++){
line += string(space, ' ');
if(extra-- > 0){
line += " ";
}
line += a[k];
}
int x = line.size();
line += string(b - x, ' ');
result.push_back(line);
}
return result;
}
};
main(){
vector<string> v = {"The", "grumpy", "wizards", "make", "toxic", "brew", "for", "the", "evil", "queen", "and", "Jack"};
Solution ob;
print_vector(ob.fullJustify(v, 13));
}["I", "love", "coding.", "here", "we", "will", "write", "some", "program"] 16输出结果
[The grumpy, wizards make, toxic brew, for the evil, queen and, Jack ,]