You are given an array items, where each items[i] = [typei, colori, namei] describes the type, color, and name of the i-th item. You are also given a rule represented by two strings, ruleKey and ruleValue.
The i-th item is said to match the rule if one of the following is true:
- ruleKey == "type" and ruleValue == typei.
- ruleKey == "color" and ruleValue == colori.
- ruleKey == "name" and ruleValue == namei.
Return the number of items that match the given rule.
Example 1
Input: items = [["phone","blue","pixel"],["computer","silver","lenovo"],["phone","gold","iphone"]], ruleKey = "color", ruleValue = "silver"
Output: 1
Explanation: There is only one item matching the given rule, which is ["computer","silver","lenovo"].
Example 2
Input: items = [["phone","blue","pixel"],["computer","silver","phone"],["phone","gold","iphone"]], ruleKey = "type", ruleValue = "phone"
Output: 2
Explanation: There are only two items matching the given rule, which are ["phone","blue","pixel"] and ["phone","gold","iphone"]. Note that the item ["computer","silver","phone"] does not match.
Constraints
- 1 <= items.length <= 1e4
- 1 <= typei.length, colori.length, namei.length, ruleValue.length <= 10
- ruleKey is equal to either "type", "color", or "name".
- All strings consist only of lowercase letters.
Solution
可以利用哈希表把输入 转换为 的下标，然后再遍历一遍 ，找出符合条件的物品数量。
class Solution {
public:
int countMatches(vector<vector<string>>& items, string ruleKey, string ruleValue) {
unordered_map<string, int> dictionary = {{"type", 0}, {"color", 1}, {"name", 2}};
int res = 0, index = dictionary[ruleKey];
for (auto &&item : items) {
if (item[index] == ruleValue) {
res++;
}
}
return res;
}
};
