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41 | 41 |
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42 | 42 | <!-- 这里可写通用的实现逻辑 -->
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43 | 43 |
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| 44 | +先用哈希表 mp 记录 list2 的每个字符串以及对应的下标。初始化最小的索引和 mi = 2000,ans 表示结果列表,初始值为空。 |
| 45 | + |
| 46 | +遍历 list1 每个字符串 v,若 v 在 mp 中,则计算两个字符串的索引和 t,并更新 ans 和 mi。 |
| 47 | + |
| 48 | +最后返回 ans 即可。 |
| 49 | + |
44 | 50 | <!-- tabs:start -->
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45 | 51 |
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46 | 52 | ### **Python3**
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47 | 53 |
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48 | 54 | <!-- 这里可写当前语言的特殊实现逻辑 -->
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49 | 55 |
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50 | 56 | ```python
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51 |
| - |
| 57 | +class Solution: |
| 58 | + def findRestaurant(self, list1: List[str], list2: List[str]) -> List[str]: |
| 59 | + ans = [] |
| 60 | + mp = {v: i for i, v in enumerate(list2)} |
| 61 | + mi = 2000 |
| 62 | + for i, v in enumerate(list1): |
| 63 | + if v in mp: |
| 64 | + t = i + mp[v] |
| 65 | + if t < mi: |
| 66 | + mi = t |
| 67 | + ans = [v] |
| 68 | + elif t == mi: |
| 69 | + ans.append(v) |
| 70 | + return ans |
52 | 71 | ```
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53 | 72 |
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54 | 73 | ### **Java**
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55 | 74 |
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56 | 75 | <!-- 这里可写当前语言的特殊实现逻辑 -->
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57 | 76 |
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58 | 77 | ```java
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| 78 | +class Solution { |
| 79 | + |
| 80 | + public String[] findRestaurant(String[] list1, String[] list2) { |
| 81 | + Map<String, Integer> mp = new HashMap<>(); |
| 82 | + for (int i = 0; i < list2.length; ++i) { |
| 83 | + mp.put(list2[i], i); |
| 84 | + } |
| 85 | + List<String> ans = new ArrayList<>(); |
| 86 | + int mi = 2000; |
| 87 | + for (int i = 0; i < list1.length; ++i) { |
| 88 | + if (mp.containsKey(list1[i])) { |
| 89 | + int t = i + mp.get(list1[i]); |
| 90 | + if (t < mi) { |
| 91 | + ans = new ArrayList<>(); |
| 92 | + ans.add(list1[i]); |
| 93 | + mi = t; |
| 94 | + } else if (t == mi) { |
| 95 | + ans.add(list1[i]); |
| 96 | + } |
| 97 | + } |
| 98 | + } |
| 99 | + return ans.toArray(new String[0]); |
| 100 | + } |
| 101 | +} |
| 102 | + |
| 103 | +``` |
| 104 | + |
| 105 | +### **C++** |
| 106 | + |
| 107 | +```cpp |
| 108 | +class Solution { |
| 109 | +public: |
| 110 | + vector<string> findRestaurant(vector<string>& list1, vector<string>& list2) { |
| 111 | + unordered_map<string, int> mp; |
| 112 | + for (int i = 0; i < list2.size(); ++i) mp[list2[i]] = i; |
| 113 | + int mi = 2000; |
| 114 | + vector<string> ans; |
| 115 | + for (int i = 0; i < list1.size(); ++i) |
| 116 | + { |
| 117 | + if (mp.count(list1[i])) |
| 118 | + { |
| 119 | + int t = i + mp[list1[i]]; |
| 120 | + if (t < mi) |
| 121 | + { |
| 122 | + ans.clear(); |
| 123 | + ans.push_back(list1[i]); |
| 124 | + mi = t; |
| 125 | + } |
| 126 | + else if (t == mi) |
| 127 | + { |
| 128 | + ans.push_back(list1[i]); |
| 129 | + } |
| 130 | + } |
| 131 | + } |
| 132 | + return ans; |
| 133 | + } |
| 134 | +}; |
| 135 | +``` |
59 | 136 |
|
| 137 | +### **Go** |
| 138 | +
|
| 139 | +```cpp |
| 140 | +func findRestaurant(list1 []string, list2 []string) []string { |
| 141 | + mp := make(map[string]int) |
| 142 | + for i, v := range list2 { |
| 143 | + mp[v] = i |
| 144 | + } |
| 145 | + mi := 2000 |
| 146 | + var ans []string |
| 147 | + for i, v := range list1 { |
| 148 | + if _, ok := mp[v]; ok { |
| 149 | + t := i + mp[v] |
| 150 | + if t < mi { |
| 151 | + ans = []string{v} |
| 152 | + mi = t |
| 153 | + } else if t == mi { |
| 154 | + ans = append(ans, v) |
| 155 | + } |
| 156 | + } |
| 157 | + } |
| 158 | + return ans |
| 159 | +} |
60 | 160 | ```
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61 | 161 |
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62 | 162 | ### **...**
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