- 模拟
先对所有字符串进行一次遍历,保证每个字符串之间有一个空格,然后对字符串分组,确定字符串的位置。
然后对每一组的字符串分配空格:遍历这一组的字符串长度,计算出剩余的空格总数,并根据空格总数分配出固有的空格和多出来的空格数量,依次添加。
注意:最后一行字符之间只能有一个空格,需要单独列出来讨论。
class Solution {
public List<String> fullJustify(String[] words, int maxWidth) {
int n = words.length, arr[] = new int[n], cnt = 0, index = 0, i = 0, j = 0, len = 0;
List<String> list = new ArrayList<>();
for(i = 0; i < n; i++){
if(i == 0){
cnt += words[i].length();
arr[i] = index;
}
else if(cnt + words[i].length()+1 <= maxWidth){
cnt += words[i].length()+1;
arr[i] = index;
}else{
cnt = words[i].length();
arr[i] = ++index;
}
}
i = 0;
while(i < n){
len = 0;
j = i;
while(j < n && arr[i] == arr[j]){
len += words[j].length();
j++;
}
if(i+1 == j){
StringBuilder sb = new StringBuilder(words[i]);
for(int k = 0; k < maxWidth-words[i].length(); k++) sb.append(" ");
list.add(sb.toString());
}else if(j == n){
StringBuilder sb = new StringBuilder(words[i]);
for(int k = i+1; k < j; k++){
sb.append(" ");
sb.append(words[k]);
}
while(sb.length() < maxWidth) sb.append(" ");
list.add(sb.toString());
}else{
StringBuilder sb = new StringBuilder(words[i]);
int space = maxWidth-len, otherspace = space % (j-i-1), certainspace = space / (j-i-1);
for(int k = i+1; k < j; k++){
if(otherspace > 0){
otherspace--;
sb.append(" ");
}
for(int l = 0; l < certainspace; l++) sb.append(" ");
sb.append(words[k]);
}
list.add(sb.toString());
}
i = j;
}
return list;
}
}
标签:cnt,int,leetcode68,空格,words,sb,字符串,对齐,文本
From: https://www.cnblogs.com/xzh-yyds/p/16615111.html