标签：splay present controls blog 相同 oat cti == size
This is a follow up of Shortest Word Distance. The only difference is now word1 could be the same as word2.
Given a list of words and two words word1 and word2, return the shortest distance between these two words in the list.
word1 and word2 may be the same and they represent two individual words in the list.
For example,
Assume that words = ["practice", "makes", "perfect", "coding", "makes"].
Given word1 = “makes”, word2 = “coding”, return 1.
Given word1 = "makes", word2 = "makes", return 3.
Note:
You may assume word1 and word2 are both in the list.
243. Shortest Word Distance 和 245. Shortest Word Distance II 的拓展，这次两个单词可能会相同。
Python:
# Time: O(n)
# Space: O(1)
class Solution:
# @param {string[]} words
# @param {string} word1
# @param {string} word2
# @return {integer}
def shortestWordDistance(self, words, word1, word2):
dist = float("inf")
is_same = (word1 == word2)
i, index1, index2 = 0, None, None
while i < len(words):
if words[i] == word1:
if is_same and index1 is not None:
dist = min(dist, abs(index1 - i))
index1 = i
elif words[i] == word2:
index2 = i
if index1 is not None and index2 is not None:
dist = min(dist, abs(index1 - index2))
i += 1
return dist
C++:
class Solution {
public:
int shortestWordDistance(vector<string>& words, string word1, string word2) {
int p1 = -1, p2 = -1, res = INT_MAX;
for (int i = 0; i < words.size(); ++i) {
int t = p1;
if (words[i] == word1) p1 = i;
if (words[i] == word2) p2 = i;
if (p1 != -1 && p2 != -1) {
if (word1 == word2 && t != -1 && t != p1) {
res = min(res, abs(t - p1));
} else if (p1 != p2) {
res = min(res, abs(p1 - p2));
}
}
}
return res;
}
};
C++:
class Solution {
public:
int shortestWordDistance(vector<string>& words, string word1, string word2) {
int p1 = words.size(), p2 = -words.size(), res = INT_MAX;
for (int i = 0; i < words.size(); ++i) {
if (words[i] == word1) p1 = word1 == word2 ? p2 : i;
if (words[i] == word2) p2 = i;
res = min(res, abs(p1 - p2));
}
return res;
}
};
C++:
class Solution {
public:
int shortestWordDistance(vector<string>& words, string word1, string word2) {
int idx = -1, res = INT_MAX;
for (int i = 0; i < words.size(); ++i) {
if (words[i] == word1 || words[i] == word2) {
if (idx != -1 && (word1 == word2 || words[i] != words[idx])) {
res = min(res, i - idx);
}
idx = i;
}
}
return res;
}
};
类似题目：
[LeetCode] 243. Shortest Word Distance 最短单词距离
[LeetCode] 245. Shortest Word Distance II 最短单词距离 II
[LeetCode] 245. Shortest Word Distance III 最短单词距离 III
标签：splay present controls blog 相同 oat cti == size
原文地址：https://www.cnblogs.com/lightwindy/p/9736293.html
