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<?xml version="1.0" encoding="utf-8"?>
<feed xmlns="http://www.w3.org/2005/Atom">
<id>https://yulan.net.cn</id>
<title>语阑gz</title>
<updated>2021-08-29T16:36:22.110Z</updated>
<generator>https://github.com/jpmonette/feed</generator>
<link rel="alternate" href="https://yulan.net.cn"/>
<link rel="self" href="https://yulan.net.cn/atom.xml"/>
<subtitle>独语斜阑</subtitle>
<logo>https://yulan.net.cn/images/avatar.png</logo>
<icon>https://yulan.net.cn/favicon.ico</icon>
<rights>All rights reserved 2021, 语阑gz</rights>
<entry>
<title type="html"><![CDATA[881. 救生艇 2021-08-26]]></title>
<id>https://yulan.net.cn/post/erlm2m_ugz5wq/</id>
<link href="https://yulan.net.cn/post/erlm2m_ugz5wq/">
</link>
<updated>2021-08-26T07:14:49.000Z</updated>
<content type="html"><![CDATA[<pre><code class="language-cpp">class Solution {
public:
int numRescueBoats(vector<int>& people, int limit) {
sort(people.begin(), people.end());
int count = 0;
vector<bool> used(people.size(), false);
for (int i = (int) people.size() - 1; i >= 0; --i) {
if (used[i])
continue;
++count;
auto pt = upper_bound(people.begin(), people.begin() + i, limit - people[i]);
if (pt == people.begin())
continue;
do {
--pt;
if (!used[pt - people.begin()]) {
used[pt - people.begin()] = true;
break;
}
} while (pt != people.begin());
}
return count;
}
};
</code></pre>
]]></content>
</entry>
<entry>
<title type="html"><![CDATA[797. 所有可能的路径 2021-08-25]]></title>
<id>https://yulan.net.cn/post/erlm2m_tmg8zs/</id>
<link href="https://yulan.net.cn/post/erlm2m_tmg8zs/">
</link>
<updated>2021-08-25T05:24:15.000Z</updated>
<content type="html"><![CDATA[<pre><code class="language-cpp">class Solution {
private:
vector<vector<int> > ans;
vector<int> path;
void DFS(int node, vector<vector<int>> &G) {
if (node == G.size() - 1) {
path.push_back(node);
ans.push_back(path);
path.pop_back();
return;
}
path.push_back(node);
for (auto i : G[node]) {
DFS(i, G);
}
path.pop_back();
}
public:
vector<vector<int>> allPathsSourceTarget(vector<vector<int>> &graph) {
DFS(0, graph);
return ans;
}
};
</code></pre>
]]></content>
</entry>
<entry>
<title type="html"><![CDATA[787. K 站中转内最便宜的航班 2021-08-24]]></title>
<id>https://yulan.net.cn/post/erlm2m_cdiz9f/</id>
<link href="https://yulan.net.cn/post/erlm2m_cdiz9f/">
</link>
<updated>2021-08-24T07:57:26.000Z</updated>
<content type="html"><![CDATA[<pre><code class="language-cpp">class Solution {
private:
const int NOT_CONNECTED = 10000000;
public:
int findCheapestPrice(int n, vector<vector<int>> &flights, int src, int dst, int k) {
vector<vector<int> > G(n, vector<int>(n, NOT_CONNECTED));
for (auto &flight : flights) {
G[flight[0]][flight[1]] = flight[2];
}
for (int i = 0; i < n; ++i)
G[i][i] = 0;
vector<vector<int> > dp(k + 2, vector<int>(n, NOT_CONNECTED));
dp[0][src] = 0;
for (int i = 1; i <= k + 1; ++i) {
for (int j = 0; j < n; ++j) {
for (int l = 0; l < n; ++l) {
dp[i][j] = min(dp[i][j], dp[i - 1][l] + G[l][j]);
}
}
}
return dp[k + 1][dst] == NOT_CONNECTED ? -1 : dp[k + 1][dst];
}
};
</code></pre>
]]></content>
</entry>
<entry>
<title type="html"><![CDATA[789. 逃脱阻碍者 2021-08-22]]></title>
<id>https://yulan.net.cn/post/erlm2m_iaa622/</id>
<link href="https://yulan.net.cn/post/erlm2m_iaa622/">
</link>
<updated>2021-08-21T23:48:17.000Z</updated>
<content type="html"><![CDATA[<pre><code class="language-cpp">class Solution {
public:
bool escapeGhosts(vector<vector<int>>& ghosts, vector<int>& target) {
int steps = abs(target[0]) + abs(target[1]);
for (auto ghost : ghosts) {
if (abs(ghost[0] - target[0]) + abs(ghost[1] - target[1]) <= steps)
return false;
}
return true;
}
};
</code></pre>
]]></content>
</entry>
<entry>
<title type="html"><![CDATA[443. 压缩字符串 2021-08-21]]></title>
<id>https://yulan.net.cn/post/erlm2m_vaxcsq/</id>
<link href="https://yulan.net.cn/post/erlm2m_vaxcsq/">
</link>
<updated>2021-08-21T05:03:29.000Z</updated>
<content type="html"><![CDATA[<pre><code class="language-cpp">class Solution {
private:
inline int digital_length(int a) {
int count = 0;
while (a > 0) {
++count;
a /= 10;
}
return count;
}
inline int put_num(vector<char> &chars, int start, int num) {
int len = digital_length(num);
for (int i = 1; i <= len; ++i) {
chars[start + len - i] = '0' + (num % 10);
num /= 10;
}
return len;
}
public:
int compress(vector<char> &chars) {
char pre;
int count;
int ans = 0;
pre = chars[0];
count = 1;
int i, j;
for (i = 1, j = 0; i < chars.size(); ++i) {
if (chars[i] == pre)
++count;
else {
chars[j++] = pre;
ans += 1;
if (count > 1) {
int len = put_num(chars, j, count);
j += len;
ans += len;
}
count = 1;
pre = chars[i];
}
}
chars[j++] = pre;
ans += 1;
if (count > 1) {
int len = put_num(chars, j, count);
ans += len;
}
return ans;
}
};
</code></pre>
]]></content>
</entry>
<entry>
<title type="html"><![CDATA[541. 反转字符串 II 2021-08-20]]></title>
<id>https://yulan.net.cn/post/erlm2m_gy90w7/</id>
<link href="https://yulan.net.cn/post/erlm2m_gy90w7/">
</link>
<updated>2021-08-19T23:31:30.000Z</updated>
<content type="html"><![CDATA[<pre><code class="language-cpp">class Solution {
public:
string reverseStr(string s, int k) {
auto it = s.begin();
while (it + 2 * k < s.end()) {
reverse(it, it + k);
it += 2 * k;
}
if (s.end() - it >= k)
reverse(it, it + k);
else
reverse(it, s.end());
return s;
}
};
</code></pre>
]]></content>
</entry>
<entry>
<title type="html"><![CDATA[345. 反转字符串中的元音字母 2021-08-19]]></title>
<id>https://yulan.net.cn/post/erlm2m_omq5ic/</id>
<link href="https://yulan.net.cn/post/erlm2m_omq5ic/">
</link>
<updated>2021-08-19T05:31:22.000Z</updated>
<content type="html"><![CDATA[<pre><code class="language-cpp">class Solution {
private:
set<char> vowels = {'a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'};
public:
string reverseVowels(string s) {
int p1 = 0, p2 = s.length() - 1;
while (p1 < p2) {
while (p1 < p2 && vowels.count(s[p1]) == 0) {
++p1;
}
while (p1 < p2 && vowels.count(s[p2]) == 0) {
--p2;
}
if (p1 < p2)
swap(s[p1], s[p2]);
++p1;
--p2;
}
return s;
}
};
</code></pre>
]]></content>
</entry>
<entry>
<title type="html"><![CDATA[552. 学生出勤记录 II 2021-08-18]]></title>
<id>https://yulan.net.cn/post/erlm2m_whqg9c/</id>
<link href="https://yulan.net.cn/post/erlm2m_whqg9c/">
</link>
<updated>2021-08-18T01:38:03.000Z</updated>
<content type="html"><![CDATA[<h2 id="dp">DP</h2>
<pre><code class="language-cpp">class Solution {
private:
static constexpr int CEIL = 1000000007;
public:
int checkRecord(int n) {
vector<vector<long long> > dp(n, vector<long long>(6, 0));
// dp[i][0~2]表示在第i天，总A数为0，连续L数为0/1/2时的总组合数，
// dp[i][3~5]表示在第i天，总A数为1，连续L数为0/1/2时的总组合数
dp[0][0] = 1; // 第一天为P
dp[0][1] = 1; // 第一天为L
dp[0][3] = 1; // 第一天为A
for (int i = 1; i < n; ++i) {
// 第i天为P
for (int j = 0; j < 3; ++j) {
dp[i][0] = (dp[i][0] + dp[i - 1][j]) % CEIL;
}
for (int j = 3; j < 6; ++j) {
dp[i][3] = (dp[i][3] + dp[i - 1][j]) % CEIL;
}
// 第i天为A
for (int j = 0; j < 3; ++j) {
dp[i][3] = (dp[i][3] + dp[i - 1][j]) % CEIL;
}
// 第i天为L
dp[i][1] = (dp[i][1] + dp[i - 1][0]) % CEIL;
dp[i][2] = (dp[i][2] + dp[i - 1][1]) % CEIL;
dp[i][4] = (dp[i][4] + dp[i - 1][3]) % CEIL;
dp[i][5] = (dp[i][5] + dp[i - 1][4]) % CEIL;
}
long long count = 0;
for (int i = 0; i < 6; ++i) {
count = (count + dp[n - 1][i]) % CEIL;
}
return count;
}
};
</code></pre>
<h2 id="优化空间复杂度">优化空间复杂度</h2>
<pre><code class="language-cpp">class Solution {
private:
static constexpr int CEIL = 1000000007;
public:
int checkRecord(int n) {
vector<long long> dp1(6, 0), dp2(6, 0);
dp1[0] = 1; // 第一天为P
dp1[1] = 1; // 第一天为L
dp1[3] = 1; // 第一天为A
for (int i = 1; i < n; ++i) {
// 第i天为P
for (int j = 0; j < 3; ++j) {
dp2[0] = (dp2[0] + dp1[j]) % CEIL;
}
for (int j = 3; j < 6; ++j) {
dp2[3] = (dp2[3] + dp1[j]) % CEIL;
}
// 第i天为A
for (int j = 0; j < 3; ++j) {
dp2[3] = (dp2[3] + dp1[j]) % CEIL;
}
// 第i天为L
dp2[1] = (dp2[1] + dp1[0]) % CEIL;
dp2[2] = (dp2[2] + dp1[1]) % CEIL;
dp2[4] = (dp2[4] + dp1[3]) % CEIL;
dp2[5] = (dp2[5] + dp1[4]) % CEIL;
dp1.swap(dp2);
fill(dp2.begin(), dp2.end(), 0);
}
long long count = 0;
for (int i = 0; i < 6; ++i) {
count = (count + dp1[i]) % CEIL;
}
return count;
}
};
</code></pre>
<p>优化后时间复杂度为</p>
<p class='katex-block'><span class="katex-display"><span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>O</mi><mo>(</mo><mi>n</mi><mo>)</mo></mrow><annotation encoding="application/x-tex">O(n)
</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord mathdefault" style="margin-right:0.02778em;">O</span><span class="mopen">(</span><span class="mord mathdefault">n</span><span class="mclose">)</span></span></span></span></span></p>
<p>空间复杂度为</p>
<p class='katex-block'><span class="katex-display"><span class="katex"><span class="katex-mathml"><math><semantics><mrow><mi>O</mi><mo>(</mo><mn>1</mn><mo>)</mo></mrow><annotation encoding="application/x-tex">O(1)
</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;"></span><span class="mord mathdefault" style="margin-right:0.02778em;">O</span><span class="mopen">(</span><span class="mord">1</span><span class="mclose">)</span></span></span></span></span></p>
]]></content>
</entry>
<entry>
<title type="html"><![CDATA[551. 学生出勤记录 I 2021-08-17]]></title>
<id>https://yulan.net.cn/post/erlm2m_hqc1dh/</id>
<link href="https://yulan.net.cn/post/erlm2m_hqc1dh/">
</link>
<updated>2021-08-17T05:24:05.000Z</updated>
<content type="html"><![CDATA[<pre><code class="language-cpp">class Solution {
public:
bool checkRecord(string s) {
int countA = 0;
int countL = 0;
for (char c : s) {
if (c == 'A') {
++countA;
if (countA == 2)
return false;
}
if (c == 'L') {
++countL;
if (countL == 3)
return false;
}
else
countL = 0;
}
return true;
}
};
</code></pre>
]]></content>
</entry>
<entry>
<title type="html"><![CDATA[526. 优美的排列 2021-08-16]]></title>
<id>https://yulan.net.cn/post/erlm2m_gvrog8/</id>
<link href="https://yulan.net.cn/post/erlm2m_gvrog8/">
</link>
<updated>2021-08-15T20:50:00.000Z</updated>
<content type="html"><![CDATA[<pre><code class="language-cpp">class Solution {
private:
vector<bool> used;
int n;
int DFS(int ind) {
if (ind == n + 1) return 1;
int count = 0;
for (int i = 1; i <= n; ++i) {
if (ind % i == 0 || i % ind == 0)
if (!used[i]) {
used[i] = true;
count += DFS(ind + 1);
used[i] = false;
}
}
return count;
}
public:
int countArrangement(int n_) {
this->n = n_;
used.resize(n + 1, false);
return DFS(1);
}
};
</code></pre>
]]></content>
</entry>
</feed>