LeetCode 30 days Challenge - Day 7
本系列将对LeetCode新推出的30天算法挑战进行总结记录,旨在记录学习成果、方便未来查阅,同时望为广大网友提供帮助。
Counting Elements
Given an integer array arr
, count element x
such that x + 1
is also in arr
.
If there’re duplicates in arr
, count them seperately.
Example 1:
Input: arr = [1,2,3]
Output: 2
Explanation: 1 and 2 are counted cause 2 and 3 are in arr.
Example 2:
Input: arr = [1,1,3,3,5,5,7,7]
Output: 0
Explanation: No numbers are counted, cause there's no 2, 4, 6, or 8 in arr.
Example 3:
Input: arr = [1,3,2,3,5,0]
Output: 3
Explanation: 0, 1 and 2 are counted cause 1, 2 and 3 are in arr.
Example 4:
Input: arr = [1,1,2,2]
Output: 2
Explanation: Two 1s are counted cause 2 is in arr.
Constraints:
1 <= arr.length <= 1000
0 <= arr[i] <= 1000
Solution
题目要求分析:给定一个字符串数组,对任一元素x,如果x+1也在数组中,记一次数。对多个相同元素x,分开计算次数。
解法:
使用unordered_map
哈希结构进行计数,即:
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建立一个由映射:
unordered_map<int, int> m;
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对题目给出的数组,遍历之,m的键值代表出现的元素,每个键对应的值记录其出现次数(因为若满足计数条件,相同元素要分开考虑,故此处记录了每个元素出现的次数)。
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遍历m,判断并更新res即可:
for (auto v : m) if (m.count(v.first + 1)) res += v.second;
(此处操作可以参考“对unordered_map进行遍历”)
(注意,此处
map.count()
不为0即“x+1也在数组中”)
int countElements(vector<int>& arr) {
int res = 0;
unordered_map<int, int> m;
for (int i : arr) m[i]++;
for (auto v : m) if (m.count(v.first + 1)) res += v.second;
return res;
}
2020/4 Karl