给你一个字符串数组 words ，每一个字符串长度都相同，令所有字符串的长度都为 n 。
每个字符串 words[i] 可以被转化为一个长度为 n - 1 的 差值整数数组 difference[i] ，其中对于 0 <= j <= n - 2 有 difference[i][j] = words[i][j+1] - words[i][j]
。注意两个字母的差值定义为它们在字母表中 位置 之差，也就是说 'a' 的位置是 0 ，'b' 的位置是 1 ，'z'
的位置是 25 。
- 比方说，字符串
"acb" 的差值整数数组是 [2 - 0, 1 - 2] = [2, -1] 。
words 中所有字符串 除了一个字符串以外 ，其他字符串的差值整数数组都相同。你需要找到那个不同的字符串。
请你返回 _ _words中 差值整数数组 不同的字符串。
示例 1：
**输入：** words = ["adc","wzy","abc"]
**输出：** "abc"
**解释：**
- "adc" 的差值整数数组是 [3 - 0, 2 - 3] = [3, -1] 。
- "wzy" 的差值整数数组是 [25 - 22, 24 - 25]= [3, -1] 。
- "abc" 的差值整数数组是 [1 - 0, 2 - 1] = [1, 1] 。
不同的数组是 [1, 1]，所以返回对应的字符串，"abc"。
示例 2：
**输入：** words = ["aaa","bob","ccc","ddd"]
**输出：** "bob"
**解释：** 除了 "bob" 的差值整数数组是 [13, -13] 以外，其他字符串的差值整数数组都是 [0, 0] 。
提示：
3 <= words.length <= 100n == words[i].length2 <= n <= 20words[i] 只含有小写英文字母。
方法一：遍历
思路与算法
注意到字符串数组 words 的长度 m 最小为 3，因此我们记 diff}_0，diff}_1，diff}_2 分别是 words}_0，words}_1，words}_2 的差值整数数组，基于此分情况讨论：
- 如果 diff}_0 = \textit{diff}_1，那么我们遍历 words}2 \sim \textit{words}{m-1，找到第一个差值整数数组不等于 diff}_0 的字符串即可。
- 否则如果 diff}_0 \neq \textit{diff}_1，那么我们只需判断 diff}_0 是否等于 diff}_2 即可。如果等于则足以说明 words}_1 是唯一一个与其他字符串的差值整数数组都不相同的字符串，因此直接返回 words}_1。反之，如果 diff}_0 不等于 diff}_2 则返回 words}_0。
代码
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| class Solution {
public:
vector<int> get(string &word) {
vector<int> diff(word.size() - 1);
for (int i = 0; i + 1 < word.size(); i++) {
diff[i] = word[i + 1] - word[i];
}
return diff;
}
string oddString(vector<string>& words) {
auto diff0 = get(words[0]);
auto diff1 = get(words[1]);
if (diff0 == diff1) {
for (int i = 2; i < words.size(); i++) {
if (diff0 != get(words[i])) {
return words[i];
}
}
}
return diff0 == get(words[2]) ? words[1] : words[0];
}
};
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[sol1-Java]1
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| class Solution {
public String oddString(String[] words) {
int[] diff0 = get(words[0]);
int[] diff1 = get(words[1]);
if (Arrays.equals(diff0, diff1)) {
for (int i = 2; i < words.length; i++) {
if (!Arrays.equals(diff0, get(words[i]))) {
return words[i];
}
}
}
return Arrays.equals(diff0, get(words[2])) ? words[1] : words[0];
}
public int[] get(String word) {
int[] diff = new int[word.length() - 1];
for (int i = 0; i + 1 < word.length(); i++) {
diff[i] = word.charAt(i + 1) - word.charAt(i);
}
return diff;
}
}
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[sol1-C#]1
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| public class Solution {
public string OddString(string[] words) {
int[] diff0 = Get(words[0]);
int[] diff1 = Get(words[1]);
if (Enumerable.SequenceEqual(diff0, diff1)) {
for (int i = 2; i < words.Length; i++) {
if (!Enumerable.SequenceEqual(diff0, Get(words[i]))) {
return words[i];
}
}
}
return Enumerable.SequenceEqual(diff0, Get(words[2])) ? words[1] : words[0];
}
public int[] Get(string word) {
int[] diff = new int[word.Length - 1];
for (int i = 0; i + 1 < word.Length; i++) {
diff[i] = word[i + 1] - word[i];
}
return diff;
}
}
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[sol1-Python3]1
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| class Solution:
def oddString(self, words: List[str]) -> str:
def get(word):
diff = [0] * (len(word) - 1)
for i in range(len(word)-1):
diff[i] = ord(word[i+1]) - ord(word[i])
return diff
diff0 = get(words[0])
diff1 = get(words[1])
if diff0 == diff1:
for i in range(2, len(words)):
if diff0 != get(words[i]):
return words[i]
return words[1] if diff0 == get(words[2]) else words[0]
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[sol1-Go]1
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| func get(word string) []int {
diff := make([]int, len(word) - 1)
for i:= 0; i + 1 < len(word); i++ {
diff[i] = int(word[i + 1]) - int(word[i])
}
return diff
}
func oddString(words []string) string {
diff0 := get(words[0])
diff1 := get(words[1])
if reflect.DeepEqual(diff0, diff1) {
for i:= 2; i < len(words); i++ {
if !reflect.DeepEqual(diff0, get(words[i])) {
return words[i]
}
}
}
if reflect.DeepEqual(diff0, get(words[2])) {
return words[1]
}
return words[0]
}
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[sol1-JavaScript]1
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| function get(word) {
let diff = new Array(word.length-1);
for (let i = 0; i + 1 < word.length; i++) {
diff[i] = word.charCodeAt(i + 1) - word.charCodeAt(i);
}
return diff.toString();
}
var oddString = function(words) {
let diff0 = get(words[0]);
let diff1 = get(words[1]);
if (diff0 === diff1) {
for (let i = 2; i < words.length; i++) {
if (diff0 != get(words[i])) {
return words[i];
}
}
}
return diff0 === get(words[2]) ? words[1] : words[0]
}
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[sol1-C]1
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| void get(const char *word, int *diff) {
int len = strlen(word);
for (int i = 0; i + 1 < len; i++) {
diff[i] = word[i + 1] - word[i];
}
}
char * oddString(char ** words, int wordsSize) {
int len = strlen(words[0]);
int diff0[len - 1];
int diff1[len - 1];
int diff[len - 1];
get(words[0], &diff0);
get(words[1], &diff1);
if (memcmp(&diff0, &diff1, sizeof(int) * (len - 1)) == 0) {
for (int i = 2; i < wordsSize; i++) {
get(words[i], &diff);
if (memcmp(&diff0, &diff, sizeof(int) * (len - 1)) != 0) {
return words[i];
}
}
}
get(words[2], &diff);
return memcmp(&diff0, &diff, sizeof(int) * (len - 1)) == 0 ? words[1] : words[0];
}
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复杂度分析
