05 August 2008
比269简单很多，给一个string作为order，检查单词是否按照这个顺序，检查相邻的字符，如果每对字符都按照顺序，最后返回true，中间发现有不对的，提前返回
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
class Solution {
public boolean isAlienSorted(String[] words, String order) {
int[] index = new int[26];
for (int i = 0; i < order.length(); ++i) {
index[order.charAt(i) - 'a'] = i;
}
search: for (int i = 0; i < words.length - 1; ++i) {
String word1 = words[i];
String word2 = words[i + 1];
// Find the first difference word1[k] != word2[k].
for (int k = 0; k < Math.min(word1.length(), word2.length()); ++k) {
if (word1.charAt(k) != word2.charAt(k)) {
// If they compare badly, it's not sorted.
if (index[word1.charAt(k) - 'a'] > index[word2.charAt(k) - 'a']) {
return false;
}
continue search;
}
}
// If we didn't find a first difference, the
// words are like ("app", "apple").
if (word1.length() > word2.length())
return false;
}
return true;
}
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
class Solution {
public boolean isAlienSorted(String[] words, String order) {
if (words == null || words.length == 0) {
return true;
}
String prev = words[0];
for (int i = 1; i < words.length; i++) {
// compare prev and curr
if(!check(prev, words[i], order)) {
return false;
}
prev = words[i];
}
return true;
}
private boolean check(String prev, String curr, String order) {
int idx1 = 0, idx2 = 0;
while (idx1 < prev.length() && idx2 < curr.length()) {
int p1 = order.indexOf(prev.charAt(idx1));
int p2 = order.indexOf(curr.charAt(idx2));
if (p1 < p2) {
return true;
} else if (p1 > p2) {
return false;
} else { // same char
idx1++;
idx2++;
}
}
if (idx1 < prev.length() && idx2 >= curr.length()) { // if prev is longer than curr
return false;
}
return true;
}
}