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行策课没事干，和晨哥窜通一下，打打周赛暖暖手。感觉好久都没写代码了。。
832 Flipping an Image
题目链接：
https://leetcode.com/contest/weekly-contest-84/problems/flipping-an-image/
class Solution {
public:
vector<vector<int>> flipAndInvertImage(vector<vector<int>>& A) {
const int size = A.size();
if (size == 0)return A;
const int length = A[0].size();
const bool odd = length % 2;
for (int i = 0; i < size; ++i)
{
for (int j = 0; j < length / 2; ++j)
{
if (A[i][j] == A[i][length - 1 - j])
{
A[i][j] ^= 1;
A[i][length - 1 - j] ^= 1;
}
}
if (odd)A[i][length / 2] ^= 1;
}
return A;
}
};833 Find And Replace in String
题目链接：
https://leetcode.com/contest/weekly-contest-84/problems/find-and-replace-in-string/
bool compare(const string& s, int size, int index, const string& source)
{
const int length = source.size();
const int len = index + length;
if (len > size)return false;
for (int i = index, j = 0; i < len; ++i, ++j)
if (s[i] != source[j])return false;
return true;
}
class Solution {
public:
string findReplaceString(string S, vector<int>& indexes, vector<string>& sources, vector<string>& targets) {
const int size = indexes.size();
const int s_size = S.size();
if (size == 0)return S;
using id = int;
vector<pair<int, id>> v(size);
for (int i = 0; i < size; ++i)
v[i] = std::make_pair(indexes[i], i);
std::sort(v.begin(), v.end(),
[](const pair<int, id>& lhs, const pair<int, id>& rhs)->bool
{return lhs.first < rhs.first; });
string ans;
if (indexes[0] != 0)
ans.append(S, 0, indexes[v[0].second]);
for (int i = 0; i < size; ++i)
{
int index = v[i].first;
id _id = v[i].second;
int next_index = (i == size - 1) ? s_size : v[i + 1].first;
if (compare(S, s_size, index, sources[_id]))
{
ans.append(targets[_id]);
int middle_index = index + sources[_id].size();
ans.append(S, middle_index, next_index - middle_index);
}
else
ans.append(S, index, next_index - index);
}
return ans;
}
};835 Image Overlap
题目链接：
https://leetcode.com/contest/weekly-contest-84/problems/image-overlap/
class Solution {
public:
int largestOverlap(vector<vector<int>>& A, vector<vector<int>>& B) {
const int size = A.size();
int largest_overlap = 0;
for (int i = 0; i < size; ++i)
for (int j = 0; j < size; ++j)
{
int count_A = 0;
int count_B = 0;
for (int p = i; p < size; ++p)
for (int q = j; q < size; ++q)
{
if (A[p - i][q - j] == 1 && B[p][q] == 1)count_A++;
if (B[p - i][q - j] == 1 && A[p][q] == 1)count_B++;
}
largest_overlap = max(largest_overlap, max(count_A, count_B));
}
return largest_overlap;
}
};834 Sum of Distances in Tree
题目链接：
https://leetcode.com/contest/weekly-contest-84/problems/sum-of-distances-in-tree/
class Solution {
private:
/*
* in dfs
* count[i] means the amount of the subtree[i]
* ans[i] means the total distances from the subtree[i] to the subroot[i].
**/
void dfs(int root, bool* visit, const vector<vector<int>>& nodes, vector<int>& count, vector<int>& ans)
{
visit[root] = true;
for (auto it = nodes[root].begin(); it != nodes[root].end(); ++it)
if (!visit[*it])
{
dfs(*it, visit, nodes, count, ans);
count[root] += count[*it];
ans[root] += count[*it] + ans[*it];
}
count[root] += 1;
}
/*
* in calculate
* ans is to be calculated as the real result.
* for each subtree, its father has been calculated properly,
* and ans[i] has a little change on ans[root],
* count[i] nodes is closer from the subroot, and (N - count[i]) is away from the subroot.
* then recur.
**/
void calculate(int root, int N, bool* visit, const vector<vector<int>>& nodes, const vector<int>& count, vector<int>& ans)
{
visit[root] = true;
for (auto it = nodes[root].begin(); it != nodes[root].end(); ++it)
if (!visit[*it])
{
ans[*it] = ans[root] - count[*it] + (N - count[*it]);
calculate(*it, N, visit, nodes, count, ans);
}
}
public:
vector<int> sumOfDistancesInTree(int N, vector<vector<int>>& edges) {
vector<vector<int>> nodes(N);
for (int i = 0; i < N - 1; ++i)
{
nodes[edges[i][0]].push_back(edges[i][1]);
nodes[edges[i][1]].push_back(edges[i][0]);
}
vector<int> ans(N, 0);
vector<int> count(N, 0); //count[i] represents the amount in the subtree[i]
bool* visit = new bool[N];
for_each(visit, visit + N, [](bool& b) {b = false; });
dfs(0, visit, nodes, count, ans);
for_each(visit, visit + N, [](bool& b) {b = false; });
calculate(0, N, visit, nodes, count, ans);
return ans;
}
};