[LeetCode]336. Palindrome Pairs
2017-01-04 22:41
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https://leetcode.com/problems/palindrome-pairs/
找出两个字符串拼接可形成回文的所有组合
遍历数组,把word和所在位置加入map,再遍历数组,内层遍历当前字符串,将字符串分隔成前后两个部分,判断这两个部分是否为回文,如果是,则将另一部分翻转,并在map中找是否存在,如果存在则这两者可以构成回文
public class Solution {
public List<List<Integer>> palindromePairs(String[] words) {
List<List<Integer>> res = new LinkedList();
HashMap<String, Integer> map = new HashMap();
for (int i = 0; i < words.length; i++) {
map.put(words[i], i);
}
for (int j = 0; j < words.length; j++) {
String word = words[j];
for (int i = 0; i <= word.length(); i++) {
String str1 = word.substring(0, i);
String str2 = word.substring(i);
if (isPalindrome(str1)) {
String str2rev = new StringBuilder(str2).reverse().toString();
if (map.containsKey(str2rev) && map.get(str2rev) != j) {
List<Integer> list = new LinkedList();
list.add(map.get(str2rev));
list.add(j);
res.add(list);
}
}
if (isPalindrome(str2)) {
String str1rev = new StringBuilder(str1).reverse().toString();
// str2.length() != 0保证"abcd","dcba"组合时不会出现重复结果
if (map.containsKey(str1rev) && map.get(str1rev) != j && str2.length() != 0) {
List<Integer> list = new LinkedList();
list.add(j);
list.add(map.get(str1rev));
res.add(list);
}
}
}
}
return res;
}
private boolean isPalindrome(String str) {
int beg = 0;
int end = str.length() - 1;
while (beg <= end) {
if (str.charAt(beg++) != str.charAt(end--)) {
return false;
}
}
return true;
}
}
找出两个字符串拼接可形成回文的所有组合
遍历数组,把word和所在位置加入map,再遍历数组,内层遍历当前字符串,将字符串分隔成前后两个部分,判断这两个部分是否为回文,如果是,则将另一部分翻转,并在map中找是否存在,如果存在则这两者可以构成回文
public class Solution {
public List<List<Integer>> palindromePairs(String[] words) {
List<List<Integer>> res = new LinkedList();
HashMap<String, Integer> map = new HashMap();
for (int i = 0; i < words.length; i++) {
map.put(words[i], i);
}
for (int j = 0; j < words.length; j++) {
String word = words[j];
for (int i = 0; i <= word.length(); i++) {
String str1 = word.substring(0, i);
String str2 = word.substring(i);
if (isPalindrome(str1)) {
String str2rev = new StringBuilder(str2).reverse().toString();
if (map.containsKey(str2rev) && map.get(str2rev) != j) {
List<Integer> list = new LinkedList();
list.add(map.get(str2rev));
list.add(j);
res.add(list);
}
}
if (isPalindrome(str2)) {
String str1rev = new StringBuilder(str1).reverse().toString();
// str2.length() != 0保证"abcd","dcba"组合时不会出现重复结果
if (map.containsKey(str1rev) && map.get(str1rev) != j && str2.length() != 0) {
List<Integer> list = new LinkedList();
list.add(j);
list.add(map.get(str1rev));
res.add(list);
}
}
}
}
return res;
}
private boolean isPalindrome(String str) {
int beg = 0;
int end = str.length() - 1;
while (beg <= end) {
if (str.charAt(beg++) != str.charAt(end--)) {
return false;
}
}
return true;
}
}
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