目录
- 题目描述:
- 示例:
- 解法:
题目描述:
给定一个整数数组 A
,返回其中元素之和可被 K
?整除的(连续、非空)子数组的数目。
示例:
输入:A = [4,5,0,-2,-3,1], K = 5
输出:7
解释:
有 7 个子数组满足其元素之和可被 K = 5 整除:
[4, 5, 0, -2, -3, 1], [5], [5, 0], [5, 0, -2, -3], [0], [0, -2, -3], [-2, -3]
提示:
1 <= A.length <= 30000
-10000 <= A[i] <= 10000
2 <= K <= 10000
解法:
class Solution {
public:
int subarraysDivByK(vector<int>& A, int K) {
vector<int> mp(K, 0);
mp[0] = 1;
int pre = 0;
int res = 0;
for(int val : A){
pre += val;
pre %= K;
if(pre < 0){
pre += K;
}
res += mp[pre];
mp[pre]++;
}
return res;
}
int subarraysDivByK2(vector<int>& A, int K) {
unordered_map<int, int> mp;
int pre = 0;
mp[0] = 1;
int res = 0;
for(int val : A){
pre += val;
for(auto it : mp){
if((pre - it.first) % K == 0){
res += it.second;
}
}
if(mp.find(pre) != mp.end()){
mp[pre]++;
}else{
mp[pre] = 1;
}
}
return res;
}
};
原文地址:https://www.cnblogs.com/zhanzq/p/11063625.html
时间: 2024-11-11 16:59:30