给你一些区域列表 regions
,每个列表的第一个区域都包含这个列表内所有其他区域。
很自然地,如果区域 X
包含区域 Y
,那么区域 X
比区域 Y
大。
给定两个区域 region1
和 region2
,找到同时包含这两个区域的 最小 区域。
如果区域列表中 r1
包含 r2
和 r3
,那么数据保证 r2
不会包含 r3
。
数据同样保证最小公共区域一定存在。
示例 1:
输入: regions = [["Earth","North America","South America"], ["North America","United States","Canada"], ["United States","New York","Boston"], ["Canada","Ontario","Quebec"], ["South America","Brazil"]], region1 = "Quebec", region2 = "New York" 输出:"North America"
提示:
2 <= regions.length <= 10^4
region1 != region2
- 所有字符串只包含英文字母和空格,且最多只有 20 个字母。
题目可转换为“求最近公共祖先”问题。
class Solution:
def findSmallestRegion(
self, regions: List[List[str]], region1: str, region2: str
) -> str:
m = {}
for region in regions:
for r in region[1:]:
m[r] = region[0]
s = set()
while m.get(region1):
s.add(region1)
region1 = m[region1]
while m.get(region2):
if region2 in s:
return region2
region2 = m[region2]
return region1
class Solution {
public String findSmallestRegion(List<List<String>> regions, String region1, String region2) {
Map<String, String> m = new HashMap<>();
for (List<String> region : regions) {
for (int i = 1; i < region.size(); ++i) {
m.put(region.get(i), region.get(0));
}
}
Set<String> s = new HashSet<>();
while (m.containsKey(region1)) {
s.add(region1);
region1 = m.get(region1);
}
while (m.containsKey(region2)) {
if (s.contains(region2)) {
return region2;
}
region2 = m.get(region2);
}
return region1;
}
}
class Solution {
public:
string findSmallestRegion(vector<vector<string>>& regions, string region1, string region2) {
unordered_map<string, string> m;
for (auto& region : regions)
for (int i = 1; i < region.size(); ++i)
m[region[i]] = region[0];
unordered_set<string> s;
while (m.count(region1)) {
s.insert(region1);
region1 = m[region1];
}
while (m.count(region2)) {
if (s.count(region2)) return region2;
region2 = m[region2];
}
return region1;
}
};
func findSmallestRegion(regions [][]string, region1 string, region2 string) string {
m := make(map[string]string)
for _, region := range regions {
for i := 1; i < len(region); i++ {
m[region[i]] = region[0]
}
}
s := make(map[string]bool)
for region1 != "" {
s[region1] = true
region1 = m[region1]
}
for region2 != "" {
if s[region2] {
return region2
}
region2 = m[region2]
}
return region1
}