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Description:
Suppose Andy and Doris want to choose a restaurant for dinner, and they both have a list of favorite restaurants represented by strings.
You need to help them find out their common interest with the least list index sum. If there is a choice tie between answers, output all of them with no order requirement. You could assume there always exists an answer.
Example 1:
Input:
["Shogun", "Tapioca Express", "Burger King", "KFC"]
["Piatti", "The Grill at Torrey Pines", "Hungry Hunter Steakhouse", "Shogun"]
Output: ["Shogun"]
Explanation: The only restaurant they both like is "Shogun".
Example 2:
Input:
["Shogun", "Tapioca Express", "Burger King", "KFC"]
["KFC", "Shogun", "Burger King"]
Output: ["Shogun"]
Explanation: The restaurant they both like and have the least index sum is "Shogun" with index sum 1 (0+1).
Note:
- The length of both lists will be in the range of [1, 1000].
- The length of strings in both lists will be in the range of [1, 30].
- The index is starting from 0 to the list length minus 1.
- No duplicates in both lists.
题意:计算两个字符串的数组的公共字符串,可能包含多个,但是要求这些公共字符串(在数组中)的下标位置和最小;
解法:这里面有一个坑是要求这些公共字符串在两个字符串数组中的下标位置和最小,那么其实结果只有两种情况:
- 返回的结果中只包含一个字符串;
- 返回的结果中包含多个字符串,这些字符串在两个字符串数组中的下标位置和相同;
利用哈希表记录第一个字符串数组中所有字符串及其下标位置后,遍历第二个字符串数组,计算过程如下
for str in list2:
if 哈希表中不存在str:
判断下一个字符串
if str为找到的第一个公共字符串:
添加str到结果字符串数组res中,并记录此时的下标位置和indexSum
elif 当前字符串的下标位置和 < indexSum:
说明res中的字符串不是下标和最小的公共串,清空res,
并添加当前字符串到res中,更新此时的下标位置和indexSum
elif 当前字符串的下标位置和 == indexSum:
说明找到了另外一个公共串,其下标位置和与res中的字符串下标位置和相同
将当前字符串添加到res中
Java
class Solution {
public String[] findRestaurant(String[] list1, String[] list2) {
StringBuilder res = new StringBuilder();
int indexSum = -1;
Map<String, Integer> map = new HashMap<>();
int ind = 0;
for (String str: list1) {
map.put(str, ind);
ind++;
}
ind = 0;
for (String str: list2) {
if (!map.containsKey(str)) {
ind++;
continue;
}
int tempSum = map.get(str) + ind;
if (indexSum == -1) {
indexSum = tempSum;
res.append(str);
} else if (tempSum < indexSum) {
indexSum = tempSum;
res.replace(0, res.length(), str);
} else if (tempSum == indexSum) {
res.append("," + str);
}
ind++;
}
return res.toString().split(",");
}
}