Given a list of strings, you could concatenate these strings together into a loop, where for each string you could choose to reverse it or not. Among all the possible loops, you need to find the lexicographically biggest string after cutting the loop, which will make the looped string into a regular one.
Specifically, to find the lexicographically biggest string, you need to experience two phases:
- Concatenate all the strings into a loop, where you can reverse some strings or not and connect them in the same order as given.
- Cut and make one breakpoint in any place of the loop, which will make the looped string into a regular one starting from the character at the cutpoint.
And your job is to find the lexicographically biggest one among all the possible regular strings.
Example:
Input: "abc", "xyz" Output: "zyxcba" Explanation: You can get the looped string "-abcxyz-", "-abczyx-", "-cbaxyz-", "-cbazyx-",
where '-' represents the looped status.
The answer string came from the fourth looped one,
where you could cut from the middle character 'a' and get "zyxcba".
Note:
- The input strings will only contain lowercase letters.
- The total length of all the strings will not over 1,000.
给定一组字符串，对于每个字符串可以逆序或者不变，将字符串按照原始顺序拼接组成一个字符串，循环每一个字符串拼接它们，可以选择循环中的一个字符串在某一个位置切开。求能得到的字典序最大的字符串。
最终结果一定是从某个字符串在某个地方切开（包括在string[0]之前切开）。其余字符串是否需要reverse则取决于reverse之后其字典序是否会增大，如果增大了就reverse，否则就保持不变，这样才能保证最后的结果是字典序最大的。对于a字符串而言，是否需要反向，以及从哪里切开，均取决于最终形成的字符串头部的字典序大小。
解法：首先将所有的字符串都处理成字典序最大的（反向或者不反向），然后连接起来。接着遍历每个字符串，试图将它以及它的reverse从每个位置切开，并测试形成的regular字符串，一旦发现其字典序大于当前的最大值，则更新当前最大值。最后返回最大值即可。
首先将res初始化为“a”，即为字典序中最小的字符串。而在测试切开位置的过程中，只有在p1[j]或者p2[j]大于res[0]的时候才更新res，这样就可以减少冗余的字符串拼接操作，从而提高运行效率。
第一步，处理原字符串，如果strs[i].reverse > strs[i], 则strs[i] = strs[i].reverse. 可以确保mid string是最优的
第二步，对于一个string array，比如 "abc", "def", "xyz", 先生成一个mid string，mid string没有最后一个string，比如"abcdef". 然后update mid string:
mid = mid.substr(str.length()) + strs[(i+n-1) % n];
这样mid string就变成 defxyz(由abcdef，去掉abc，加上xyz)，xyzabc.
然后扫原string array中的每一个string，生成相应的截取result。比如对于abc，我们有mid string = defxyz, 所以组合是：
正序：abc-defxyz, bc-defxyz-a, c-defxyz-ab, defxyz-abc
反序：cba-defxyz, ba-defxyz-c, a-defxyz-cb, defxyz-cba,
然后所有循环中挑出全局最大的.
参考 For every given string, we replace the string with the lexicographically larger string out of the original string and the reversed one. After this, we pick up every new string(chosen as the string on which the cuts will be applied), and apply a cut at all the positions of the currently picked string and form the full concantenated string keeping the rest of the newly formed strings intact. We also reverse the current string and follow the same process. While doing this, we keep a track of the largest lexicographic string found so far.
大厂题：阿里巴巴
Java:
public class Solution {
public String splitLoopedString(String[] strs) {
// 先各自翻转一下
for (int i = 0; i < strs.length; i++) {
String rev = new StringBuilder(strs[i]).reverse().toString();
if (strs[i].compareTo(rev) < 0)
strs[i] = rev;
}
String res = "";
// 尝试every str作为那个被cut的
for (int i = 0; i < strs.length; i++) {
String rev = new StringBuilder(strs[i]).reverse().toString();
// 比较其 翻转 和 没翻转
for (String st: new String[] {strs[i], rev}) {
// try every k 作为分界点
for (int k = 0; k < st.length(); k++) {
// 拼接其他strs到t
StringBuilder t = new StringBuilder(st.substring(k));
for (int j = i + 1; j < strs.length; j++)
t.append(strs[j]);
for (int j = 0; j < i; j++)
t.append(strs[j]);
t.append(st.substring(0, k));
// 看t是否最大
if (t.toString().compareTo(res) > 0)
res = t.toString();
}
}
}
return res;
}
}
Java: DFS
public class Solution {
String res = "";
public String splitLoopedString(String[] strs) {
dfs(strs, "", 0, strs.length);
return res;
}
public void dfs(String[] strs, String s, int i, int n) {
if (i < n) {
dfs(strs, s + strs[i], i + 1, n);
dfs(strs, s + new StringBuffer(strs[i]).reverse().toString(), i + 1, n);
} else {
for (int j = 0; j < s.length(); j++) {
String t = s.substring(j) + s.substring(0, j);
if (t.compareTo(res) > 0)
res = t;
}
}
}
}
Java: BFS
public class Solution {
public String splitLoopedString(String[] strs) {
Queue < String > queue = new LinkedList < > ();
String res = "";
int i = 0, j = 0;
queue.add("");
while (i < strs.length) {
String t = queue.remove();
queue.add(t + strs[i]);
queue.add(t + new StringBuffer(strs[i]).reverse().toString());
j++;
if (j == 1 << i) {
i++;
j = 0;
}
}
while (!queue.isEmpty()) {
String t = queue.remove();
for (int k = 0; k < t.length(); k++) {
String t1 = t.substring(k) + t.substring(0, k);
if (t1.compareTo(res) > 0)
res = t1;
}
}
return res;
}
}
Java: Optimized Solution
public class Solution {
public String splitLoopedString(String[] strs) {
for (int i = 0; i < strs.length; i++) {
String rev = new StringBuilder(strs[i]).reverse().toString();
if (strs[i].compareTo(rev) < 0)
strs[i] = rev;
}
String res = "";
for (int i = 0; i < strs.length; i++) {
String rev = new StringBuilder(strs[i]).reverse().toString();
for (String st: new String[] {strs[i], rev}) {
for (int k = 0; k < st.length(); k++) {
StringBuilder t = new StringBuilder(st.substring(k));
for (int j = i + 1; j < strs.length; j++)
t.append(strs[j]);
for (int j = 0; j < i; j++)
t.append(strs[j]);
t.append(st.substring(0, k));
if (t.toString().compareTo(res) > 0)
res = t.toString();
}
}
}
return res;
}
}
Python:
class Solution(object):
def splitLoopedString(self, strs):
"""
:type strs: List[str]
:rtype: str
"""
strs = [max(s, s[::-1]) for s in strs]
ans = ''
for i, st in enumerate(strs):
left, right = ''.join(strs[:i]), ''.join(strs[i+1:])
for s in (st, st[::-1]):
for j in range(len(s)):
ans = max(ans, s[j:] + right + left + s[:j])
return ans
C++:
class Solution {
public:
string splitLoopedString(vector<string>& strs) {
if(strs.empty()) return "";
else if(strs.size() == 1) return max(strs[0], string(strs[0].rbegin(), strs[0].rend()));
string all = "";
int n = strs.size();
for(int i=0; i<n; i++){
string temp = string(strs[i].rbegin(), strs[i].rend());
if(temp > strs[i]) strs[i] = temp;
}
for(int i=0; i<n-1; i++){
all += strs[i];
}
string result = all + strs[n-1];
for(int i=0; i<n; i++){
string str = strs[i], rev = string(strs[i].rbegin(), strs[i].rend());
all = all.substr(str.length()) + strs[(i+n-1) % n];
for(int j=0; j<=str.length(); j++){
string s1 = str.substr(j) + all + str.substr(0, j), s2 = rev.substr(j) + all + rev.substr(0, j);
if(s1 >= s2 && s1 > result) result = s1;
else if(s2 >= s1 && s2 > result) result = s2;
}
}
return result;
}
};
C++:
class Solution {
public:
string splitLoopedString(vector<string>& strs) {
if (strs.empty()) return "";
string s = "", res = "a";
int n = strs.size(), cur = 0;
for (string str : strs) {
string t = string(str.rbegin(), str.rend());
s += str > t ? str : t;
}
for (int i = 0; i < n; ++i) {
string t1 = strs[i], t2 = string(t1.rbegin(), t1.rend());
string mid = s.substr(cur + t1.size()) + s.substr(0, cur);
for (int j = 0; j < strs[i].size(); ++j) {
if (t1[j] >= res[0]) res = max(res, t1.substr(j) + mid + t1.substr(0, j));
if (t2[j] >= res[0]) res = max(res, t2.substr(j) + mid + t2.substr(0, j));
}
cur += strs[i].size();
}
return res;
}
};