Merge pull request #637 from LetMeFly666/632

LetMeFly666 · web-flow · commit 37a4490d3d63 · 2024-11-24T19:10:45.000+08:00
添加问题“632.最小区间”的代码和题解
diff --git a/Codes/0632-smallest-range-covering-elements-from-k-lists.cpp b/Codes/0632-smallest-range-covering-elements-from-k-lists.cpp
@@ -0,0 +1,174 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2024-11-24 17:37:00
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2024-11-24 18:31:53
+ */
+#ifdef _WIN32
+#include "_[1,2]toVector.h"
+#endif
+
+#ifdef FirstTry  // TLE(92 / 94 个通过的测试用例)  // 最后可能全部移动的最后一组的指针，但是前面的元素还要全部再判断并continue一次
+/*
+4 10 15
+0 9 12
+
+0-4
+4-9
+9-10
+10-12
+12-15
+
+min: 9-10
+*/
+class Solution {
+private:
+    void update(vector<vector<int>>& nums, vector<int>& loc, int& m, int& M) {
+        m = 1e6, M = -1e6;
+        for (int i = 0; i < nums.size(); i++) {
+            m = min(m, nums[i][loc[i]]);
+            M = max(M, nums[i][loc[i]]);
+        }
+    }
+public:
+    vector<int> smallestRange(vector<vector<int>>& nums) {
+        // init
+        vector<int> loc(nums.size());
+        int m, M;
+        update(nums, loc, m, M);
+        int minDiff = M - m;
+        // begin
+        while (true) {
+            int moveWhich = -1, minChange = 1e6;
+            for (int i = 0; i < nums.size(); i++) {
+                if (loc[i] == nums[i].size() - 1) {
+                    continue;
+                }
+                if (nums[i][loc[i] + 1] - M < minChange) {
+                    moveWhich = i;
+                    minChange = nums[i][loc[i] + 1] - M;
+                }
+            }
+            if (moveWhich == -1) {
+                break;
+            }
+            loc[moveWhich]++;
+            int thism, thisM;
+            update(nums, loc, thism, thisM);
+            if (thisM - thism < minDiff) {
+                minDiff = thisM - thism;
+                m = thism, M = thisM;
+            }
+        }
+        return {m, M};
+    }
+};
+#else  // FirstTry
+#ifdef SecondTry  // TLE  // 时间瓶颈卡在获取m和M上
+class Solution {
+private:
+    void updateMm(vector<vector<int>>& nums, vector<int>& loc, int& m, int& M) {
+        m = 1e6, M = -1e6;
+        for (int i = 0; i < nums.size(); i++) {
+            m = min(m, nums[i][loc[i]]);
+            M = max(M, nums[i][loc[i]]);
+        }
+    }
+public:
+    vector<int> smallestRange(vector<vector<int>>& nums) {
+        vector<int> loc(nums.size());
+        int m, M;
+        updateMm(nums, loc, m, M);
+        int minDiff = M - m;
+        priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> pq;
+        for (int i = 0; i < nums.size(); i++) {
+            for (int j = 1; j < nums[i].size(); j++) {
+                pq.push({nums[i][j], i});
+            }
+        }
+        while (pq.size()) {
+            auto&& [val, index] = pq.top();
+            loc[index]++;
+            pq.pop();
+            int thism, thisM;
+            updateMm(nums, loc, thism, thisM);
+            if (thisM - thism < minDiff) {
+                minDiff = thisM - thism;
+                m = thism, M = thisM;                
+            }
+        }
+        return {m, M};
+    }
+};
+#else  // SecondTry
+#ifdef ThirdTry  // WA - 最终会收敛到[最大数, 最大数]
+class Solution {
+public:
+    vector<int> smallestRange(vector<vector<int>>& nums) {
+        vector<int> loc(nums.size());
+        // lambda [](const int& x, const int& y)->bool cmp = xxx
+        auto cmp = [&nums, &loc](const int& x, const int& y) {
+            return nums[x][loc[x]] > nums[y][loc[y]];
+        };
+        priority_queue<int, vector<int>, decltype(cmp)> pq(cmp);
+        int m = -1e8, M = -1e6;
+        for (int i = 0; i < nums.size(); i++) {
+            pq.push(i);
+            M = max(M, nums[i][0]);
+        }
+        int nowm, nowM = M;
+        while (pq.size()) {
+            int index = pq.top();
+            pq.pop();
+            nowm = nums[index][loc[index]];
+            loc[index]++;
+            if (nowM - nowm < M - m) {
+                M = nowM;
+                m = nowm;
+            }
+            if (loc[index] < nums[index].size()) {
+                nowM = max(nowM, nums[index][loc[index]]);
+                pq.push(index);
+            }
+            // printf("m = %d, M = %d, nowm = %d, nowM = %d\n", m, M, nowm, nowM);  //*************
+        }
+        return {m, M};
+    }
+};
+#else  // ThirdTry
+// FourthTry
+class Solution {
+public:
+    vector<int> smallestRange(vector<vector<int>>& nums) {
+        vector<int> loc(nums.size());
+        auto cmp = [&nums, &loc](const int& x, const int& y) {
+            return nums[x][loc[x]] > nums[y][loc[y]];
+        };
+        priority_queue<int, vector<int>, decltype(cmp)> pq(cmp);
+        int m = -1e8, M = -1e6;
+        for (int i = 0; i < nums.size(); i++) {
+            pq.push(i);
+            M = max(M, nums[i][0]);
+        }
+        int nowm, nowM = M;
+        while (pq.size()) {
+            int index = pq.top();
+            pq.pop();
+            nowm = nums[index][loc[index]];
+            loc[index]++;
+            if (nowM - nowm < M - m) {
+                M = nowM;
+                m = nowm;
+            }
+            if (loc[index] == nums[index].size()) {
+                break;
+            }
+            nowM = max(nowM, nums[index][loc[index]]);
+            pq.push(index);
+        }
+        return {m, M};
+    }
+};
+#endif  // ThirdTry
+#endif  // SecondTry
+#endif  // FirstTry
diff --git a/README.md b/README.md
@@ -227,6 +227,7 @@
 |0617.合并二叉树|简单|<a href="https://leetcode.cn/problems/merge-two-binary-trees/solutions/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/08/14/LeetCode%200617.%E5%90%88%E5%B9%B6%E4%BA%8C%E5%8F%89%E6%A0%91/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/132268393" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/merge-two-binary-trees/solutions/2387253/letmefly-617he-bing-er-cha-shu-by-tisfy-3f89/" target="_blank">LeetCode题解</a>|
 |0623.在二叉树中增加一行|中等|<a href="https://leetcode.cn/problems/add-one-row-to-tree/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2022/08/05/LeetCode%200623.%E5%9C%A8%E4%BA%8C%E5%8F%89%E6%A0%91%E4%B8%AD%E5%A2%9E%E5%8A%A0%E4%B8%80%E8%A1%8C/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/126179967" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/add-one-row-to-tree/solution/letmefly-623zai-er-cha-shu-zhong-zeng-ji-ajg0/" target="_blank">LeetCode题解</a>|
 |0630.课程表III|困难|<a href="https://leetcode.cn/problems/course-schedule-iii/solutions/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/09/11/LeetCode%200630.%E8%AF%BE%E7%A8%8B%E8%A1%A8III/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/132806660" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/course-schedule-iii/solutions/2437116/letmefly-630ke-cheng-biao-iiitan-xin-you-ah0l/" target="_blank">LeetCode题解</a>|
+|0632.最小区间|困难|<a href="https://leetcode.cn/problems/smallest-range-covering-elements-from-k-lists/solutions/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/11/24/LeetCode%200632.%E6%9C%80%E5%B0%8F%E5%8C%BA%E9%97%B4/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/144011635" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/smallest-range-covering-elements-from-k-lists/solutions/2999312/letmefly-632zui-xiao-qu-jian-you-xian-du-bct2/" target="_blank">LeetCode题解</a>|
 |0633.平方数之和|中等|<a href="https://leetcode.cn/problems/sum-of-square-numbers/solutions/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/11/04/LeetCode%200633.%E5%B9%B3%E6%96%B9%E6%95%B0%E4%B9%8B%E5%92%8C/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/143495591" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/sum-of-square-numbers/solutions/2977304/letmefly-633ping-fang-shu-zhi-he-mo-ni-b-2und/" target="_blank">LeetCode题解</a>|
 |0638.大礼包|中等|<a href="https://leetcode.cn/problems/shopping-offers/solutions/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/11/03/LeetCode%200638.%E5%A4%A7%E7%A4%BC%E5%8C%85/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/143465993" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/shopping-offers/solutions/2975699/letmefly-638da-li-bao-ji-yi-hua-sou-suo-qyhoh/" target="_blank">LeetCode题解</a>|
 |0640.求解方程|中等|<a href="https://leetcode.cn/problems/solve-the-equation/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2022/08/10/LeetCode%200640.%E6%B1%82%E8%A7%A3%E6%96%B9%E7%A8%8B/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/126262898" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/solve-the-equation/solution/letmefly-640qiu-jie-fang-cheng-guo-ji-ti-xbzp/" target="_blank">LeetCode题解</a>|
diff --git a/Solutions/LeetCode 0632.最小区间.md b/Solutions/LeetCode 0632.最小区间.md
@@ -0,0 +1,109 @@
+---
+title: 632.最小区间
+date: 2024-11-24 18:36:13
+tags: [题解, LeetCode, 困难, 贪心, 数组, 哈希表, 排序, 滑动窗口, 堆（优先队列）]
+---
+
+# 【LetMeFly】632.最小区间：优先队列
+
+力扣题目链接：[https://leetcode.cn/problems/smallest-range-covering-elements-from-k-lists/](https://leetcode.cn/problems/smallest-range-covering-elements-from-k-lists/)
+
+<p>你有&nbsp;<code>k</code>&nbsp;个 <strong>非递减排列</strong> 的整数列表。找到一个 <strong>最小 </strong>区间，使得&nbsp;<code>k</code>&nbsp;个列表中的每个列表至少有一个数包含在其中。</p>
+
+<p>我们定义如果&nbsp;<code>b-a &lt; d-c</code>&nbsp;或者在&nbsp;<code>b-a == d-c</code>&nbsp;时&nbsp;<code>a &lt; c</code>，则区间 <code>[a,b]</code> 比 <code>[c,d]</code> 小。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre>
+<strong>输入：</strong>nums = [[4,10,15,24,26], [0,9,12,20], [5,18,22,30]]
+<strong>输出：</strong>[20,24]
+<strong>解释：</strong> 
+列表 1：[4, 10, 15, 24, 26]，24 在区间 [20,24] 中。
+列表 2：[0, 9, 12, 20]，20 在区间 [20,24] 中。
+列表 3：[5, 18, 22, 30]，22 在区间 [20,24] 中。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<strong>输入：</strong>nums = [[1,2,3],[1,2,3],[1,2,3]]
+<strong>输出：</strong>[1,1]
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>nums.length == k</code></li>
+	<li><code>1 &lt;= k &lt;= 3500</code></li>
+	<li><code>1 &lt;= nums[i].length &lt;= 50</code></li>
+	<li><code>-10<sup>5</sup> &lt;= nums[i][j] &lt;= 10<sup>5</sup></code></li>
+	<li><code>nums[i]</code> 按非递减顺序排列</li>
+</ul>
+
+<p>&nbsp;</p>
+
+
+    
+## 解题方法：优先队列
+
+使用一个优先队列，每个“整数列表”放一个元素到优先队列中，优先队列以“列表元素最小”为最优先。
+
+优先队列中存放的，是每个列表本次要覆盖的元素。
+
+每次从优先队列中取出一个元素：
+
+> 那么这次方案(取出之前)的最小值就是取出的这个元素，最大值我们使用一个值记录并在入队时候更新。
+>
+> 更新最佳方案：如果当前方案优于之前的最佳方案，就更新最佳方案为这个方案。
+>
+> 新元素入队：如果出队元素所在列表还有新元素，则下一个元素入队，并记得更新“最大值”；否则结束循环。
+
++ 时间复杂度$O(nk\log k)$
++ 空间复杂度$O(k)$
+
+### AC代码
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    vector<int> smallestRange(vector<vector<int>>& nums) {
+        vector<int> loc(nums.size());
+        auto cmp = [&nums, &loc](const int& x, const int& y) {
+            return nums[x][loc[x]] > nums[y][loc[y]];
+        };
+        priority_queue<int, vector<int>, decltype(cmp)> pq(cmp);
+        int m = -1e8, M = -1e6;
+        for (int i = 0; i < nums.size(); i++) {
+            pq.push(i);
+            M = max(M, nums[i][0]);
+        }
+        int nowm, nowM = M;
+        while (pq.size()) {
+            int index = pq.top();
+            pq.pop();
+            nowm = nums[index][loc[index]];
+            loc[index]++;
+            if (nowM - nowm < M - m) {
+                M = nowM;
+                m = nowm;
+            }
+            if (loc[index] == nums[index].size()) {
+                break;
+            }
+            nowM = max(nowM, nums[index][loc[index]]);
+            pq.push(index);
+        }
+        return {m, M};
+    }
+};
+```
+
+> 同步发文于CSDN和我的[个人博客](https://blog.letmefly.xyz/)，原创不易，转载经作者同意后请附上[原文链接](https://blog.letmefly.xyz/2024/11/24/LeetCode%200632.%E6%9C%80%E5%B0%8F%E5%8C%BA%E9%97%B4/)哦~
+>
+> Tisfy：[https://letmefly.blog.csdn.net/article/details/144011635](https://letmefly.blog.csdn.net/article/details/144011635)
diff --git a/Solutions/Other-English-LearningNotes-SomeWords.md b/Solutions/Other-English-LearningNotes-SomeWords.md
@@ -820,6 +820,9 @@ tags: [其他, 知识, 英语, Notes]
 |hydraulic|adj. 水力的，水压的，液压驱动的|
 |curtail|v. 缩短，限制，减缩|
 |colonel|n. (陆军/海军陆战队/美国空军)上校|
+|||
+|walnut|n. 核桃，胡桃，胡桃木，胡桃色|
+|smack|v. 用巴掌打，掴，啪的一声使劲放(或扔、甩等)，使劲碰(或撞)<br/>n. 打巴掌，(打出的)一拳，啪的一声；海洛因<br/>adv. 恰好，直接，不偏不倚地，猛地|
 
 <p class="wordCounts">单词收录总数</p>
 
diff --git a/Solutions/Other-Japanese-LearningNotes.md b/Solutions/Other-Japanese-LearningNotes.md
@@ -1074,6 +1074,7 @@ xx型
 |もう夏になりました。<br/>已经夏天了。|
 |外国に行ったことがありますか？<br/>去过外国吗？|
 |明日映画を見るつもりです。<br/>明天打算去看电影。|
+|今年はカナダに行くつもりです。<br/>今年打算去加拿大。|
 |このアニメを<font color="#28bea0">見たこと</font>があります。<br/>看过这个动漫。|
 |段々暖かく<font color="#ce82ff">なり</font>ました。<br/>越来越暖和了。|
 |その<ruby>後<rt>go</rt></ruby>、映画館に行きました。<br/>那之后去了电影院。|
diff --git a/history.del-视频压制.py b/history.del-视频压制.py
diff --git a/tryGoPy/temp.多次点击.html b/tryGoPy/temp.多次点击.html
