第 119 场双周赛
找到两个数组中的公共元素
模拟
class Solution {
public:
vector<int> findIntersectionValues(vector<int>& nums1, vector<int>& nums2) {
unordered_map<int,int> mp1 , mp2;
vector<int>res(2);
for(auto x : nums1){
mp1[x] ++;
}
for(auto x : nums2){
mp2[x] ++;
}
int k = 0;
for(auto x : nums1){
if(mp2[x] > 0)k ++;
}
res[0] = k;
k = 0;
for(auto x : nums2){
if(mp1[x] > 0)k ++;
}
res[1] = k;
return res;
}
};
消除相邻近似相等字符
贪心，一个变了下一个肯定不用变
class Solution {
public:
int removeAlmostEqualCharacters(string word) {
int n = word.size() , ans = 0;
for(int i = 1 ; i < n ; i ++)if(abs(word[i] - word[i-1]) <= 1)ans ++,i ++;
return ans;
}
};
最多 K 个重复元素的最长子数组
滑动窗扣维持k个
class Solution {
public:
int maxSubarrayLength(vector<int>& nums, int k) {
int n = nums.size();
unordered_map<int,int>m;
queue<int>q;
int ans = 0;
for(int i = 0 ; i < n ; i ++){
m[nums[i]] ++;
q.push(nums[i]);
while(m[nums[i]] > k){
int t = q.front();
q.pop();
m[t] --;
}
int l = q.size();
ans = max(ans , l);
}
return ans;
}
};
找到最大非递减数组的长度
二进制枚举+Floyd
class Solution {
public:
static const int N = 11;
int d[N][N];
void init(){
for(int i = 0 ; i < N ; i ++){
for(int j = 0 ; j < N ; j ++){
if(i == j)d[i][j] = 0;
else d[i][j] = 99999;
}
}
}
int numberOfSets(int n, int md, vector<vector<int>>& roads) {
int ans = 1 , cnt = roads.size();
for(int i = 1 ; i <= ((1 << n) - 1) ; i ++){
//枚举
//建图
init();
vector<int> c;
vector<int> nt(15);
for(int j = 0 ; j <= 10 ; j ++){
if(i >> j & 1)c.push_back(j);
}
for(auto x : c)nt[x] = 1;
for(int j = 0 ; j < cnt ; j ++){
if(nt[roads[j][0]] == 1 && nt[roads[j][1]] == 1){
d[roads[j][0]][roads[j][1]] = min(d[roads[j][0]][roads[j][1]],roads[j][2]);
d[roads[j][1]][roads[j][0]] = min(d[roads[j][1]][roads[j][0]],roads[j][2]);
}
}
function<bool()> check = [&]() -> bool{
for (int kk = 0; kk < n; kk ++ )
for (int ii = 0; ii < n; ii ++ )
for (int jj = 0; jj < n; jj ++ )
d[ii][jj] = min(d[ii][jj], d[ii][kk] + d[kk][jj]);
for(int a = 0 ; a < c.size() ; a ++){
for(int b = 0 ; b < c.size() ; b ++){
if(d[c[a]][c[b]] > md){
return false;
}
}
}
return true;
};
if(check())ans ++;
}
return ans;
}
};
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