From 0a9653df8457535038888a57e05ba873109eccce Mon Sep 17 00:00:00 2001 From: Yukun J Date: Wed, 24 Aug 2022 11:22:43 -0400 Subject: [PATCH 1/4] =?UTF-8?q?Update:=200416=20=E4=BD=BF=E7=94=A8STL?= =?UTF-8?q?=E6=B1=82=E5=92=8C?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0416.分割等和子集.md | 7 ++----- 1 file changed, 2 insertions(+), 5 deletions(-) diff --git a/problems/0416.分割等和子集.md b/problems/0416.分割等和子集.md index 03eae8ef..9c90ea27 100644 --- a/problems/0416.分割等和子集.md +++ b/problems/0416.分割等和子集.md @@ -143,15 +143,12 @@ dp[j]的数值一定是小于等于j的。 class Solution { public: bool canPartition(vector& nums) { - int sum = 0; - + // 使用标准库函数 便捷求和 + int sum = accumulate(nums.begin(), nums.end(), 0); // dp[i]中的i表示背包内总和 // 题目中说：每个数组中的元素不会超过 100，数组的大小不会超过 200 // 总和不会大于20000，背包最大只需要其中一半，所以10001大小就可以了 vector dp(10001, 0); - for (int i = 0; i < nums.size(); i++) { - sum += nums[i]; - } if (sum % 2 == 1) return false; int target = sum / 2; From 55648dcb7905550ce30081295deee4de858fef35 Mon Sep 17 00:00:00 2001 From: SwaggyP <1352164869@qq.com> Date: Sun, 28 Aug 2022 10:26:45 +0800 Subject: [PATCH 2/4] =?UTF-8?q?0127.=E5=8D=95=E8=AF=8D=E6=8E=A5=E9=BE=99?= =?UTF-8?q?=20=E5=A2=9E=E5=8A=A0go=E8=AF=AD=E8=A8=80=E7=9A=84=E8=A7=A3?= =?UTF-8?q?=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 0127.单词接龙 增加go语言的解法 --- problems/0127.单词接龙.md | 51 +++++++++++++++++++++++++++++++++++ 1 file changed, 51 insertions(+) diff --git a/problems/0127.单词接龙.md b/problems/0127.单词接龙.md index f1c6f182..6d719691 100644 --- a/problems/0127.单词接龙.md +++ b/problems/0127.单词接龙.md @@ -158,6 +158,57 @@ class Solution: return 0 ``` ## Go +```go +func ladderLength(beginWord string, endWord string, wordList []string) int { + wordMap, que, depth := getWordMap(wordList, beginWord), []string{beginWord}, 0 + for len(que) > 0 { + depth++ + qLen := len(que) // 单词的长度 + for i := 0; i < qLen; i++ { + word := que[0] + que = que[1:] // 首位单词出队 + candidates := getCandidates(word) + for _, candidate := range candidates { + if _, exist := wordMap[candidate]; exist { // 用生成的结果集去查询 + if candidate == endWord { + return depth + 1 + } + delete(wordMap, candidate) // 删除集合中的用过的结果 + que = append(que, candidate) + } + } + } + } + return 0 +} + + +// 获取单词Map为后续的查询增加速度 +func getWordMap(wordList []string, beginWord string) map[string]int { + wordMap := make(map[string]int) + for i, word := range wordList { + if _, exist := wordMap[word]; !exist { + if word != beginWord { + wordMap[word] = i + } + } + } + return wordMap +} + +// 用26个英文字母分别替换掉各个位置的字母，生成一个结果集 +func getCandidates(word string) []string { + var res []string + for i := 0; i < 26; i++ { + for j := 0; j < len(word); j++ { + if word[j] != byte(int('a')+i) { + res = append(res, word[:j]+string(int('a')+i)+word[j+1:]) + } + } + } + return res +} +``` ## JavaScript ```javascript From bc8030b45a7e2a6cdefe8944cb624d4fa1f7d1a1 Mon Sep 17 00:00:00 2001 From: = <=> Date: Sun, 28 Aug 2022 14:31:18 +0800 Subject: [PATCH 3/4] =?UTF-8?q?Update=200347.=E5=89=8DK=E4=B8=AA=E9=AB=98?= =?UTF-8?q?=E9=A2=91=E5=85=83=E7=B4=A0.md=EF=BC=8C=E6=96=B0=E5=A2=9ECompar?= =?UTF-8?q?ator=E6=8E=A5=E5=8F=A3=E8=AF=B4=E6=98=8E=E5=8F=8A=E5=A4=A7?= =?UTF-8?q?=E9=A1=B6=E5=A0=86=E3=80=81=E5=B0=8F=E9=A1=B6=E5=A0=86=E4=B8=A4?= =?UTF-8?q?=E7=A7=8Djava=E5=AE=9E=E7=8E=B0?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0347.前K个高频元素.md | 59 +++++++++++++++++++++------- 1 file changed, 45 insertions(+), 14 deletions(-) diff --git a/problems/0347.前K个高频元素.md b/problems/0347.前K个高频元素.md index d4059b9b..8256e629 100644 --- a/problems/0347.前K个高频元素.md +++ b/problems/0347.前K个高频元素.md @@ -140,24 +140,55 @@ public: Java： ```java +/*Comparator接口说明: + * 返回负数，形参中第一个参数排在前面；返回正数，形参中第二个参数排在前面 + * 对于队列：排在前面意味着往队头靠 + * 对于堆（使用PriorityQueue实现）：从队头到队尾按从小到大排就是最小堆（小顶堆）， + * 从队头到队尾按从大到小排就是最大堆（大顶堆）--->队头元素相当于堆的根节点 + * */ class Solution { - public int[] topKFrequent(int[] nums, int k) { - int[] result = new int[k]; - HashMap map = new HashMap<>(); - for (int num : nums) { - map.put(num, map.getOrDefault(num, 0) + 1); + //解法1：基于大顶堆实现 + public int[] topKFrequent1(int[] nums, int k) { + Map map = new HashMap<>();//key为数组元素值,val为对应出现次数 + for(int num:nums){ + map.put(num,map.getOrDefault(num,0)+1); } - - Set> entries = map.entrySet(); - // 根据map的value值，构建于一个大顶堆（o1 - o2: 小顶堆， o2 - o1 : 大顶堆） - PriorityQueue> queue = new PriorityQueue<>((o1, o2) -> o2.getValue() - o1.getValue()); - for (Map.Entry entry : entries) { - queue.offer(entry); + //在优先队列中存储二元组(num,cnt),cnt表示元素值num在数组中的出现次数 + //出现次数按从队头到队尾的顺序是从大到小排,出现次数最多的在队头(相当于大顶堆) + PriorityQueue pq = new PriorityQueue<>((pair1, pair2)->pair2[1]-pair1[1]); + for(Map.Entry entry:map.entrySet()){//大顶堆需要对所有元素进行排序 + pq.add(new int[]{entry.getKey(),entry.getValue()}); } - for (int i = k - 1; i >= 0; i--) { - result[i] = queue.poll().getKey(); + int[] ans = new int[k]; + for(int i=0;i map = new HashMap<>();//key为数组元素值,val为对应出现次数 + for(int num:nums){ + map.put(num,map.getOrDefault(num,0)+1); + } + //在优先队列中存储二元组(num,cnt),cnt表示元素值num在数组中的出现次数 + //出现次数按从队头到队尾的顺序是从小到大排,出现次数最低的在队头(相当于小顶堆) + PriorityQueue pq = new PriorityQueue<>((pair1,pair2)->pair1[1]-pair2[1]); + for(Map.Entry entry:map.entrySet()){//小顶堆只需要维持k个元素有序 + if(pq.size()pq.peek()[1]){//当前元素出现次数大于小顶堆的根结点(这k个元素中出现次数最少的那个) + pq.poll();//弹出队头(小顶堆的根结点),即把堆里出现次数最少的那个删除,留下的就是出现次数多的了 + pq.add(new int[]{entry.getKey(),entry.getValue()}); + } + } + } + int[] ans = new int[k]; + for(int i=k-1;i>=0;i--){//依次弹出小顶堆,先弹出的是堆的根,出现次数少,后面弹出的出现次数多 + ans[i] = pq.poll()[0]; + } + return ans; } } ``` From a275c2cc553ad35e50d736e0630947a5bdf2d13e Mon Sep 17 00:00:00 2001 From: YukunJ Date: Tue, 30 Aug 2022 21:51:30 -0400 Subject: [PATCH 4/4] =?UTF-8?q?Update:=20=E6=B3=A8=E9=87=8A=E5=A2=9E?= =?UTF-8?q?=E5=8A=A0=E4=BD=BF=E7=94=A8STL=E6=B1=82=E5=92=8C=E7=9A=84?= =?UTF-8?q?=E7=AE=80=E4=BE=BF=E5=B0=8F=E8=B4=B4=E5=A3=AB?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0416.分割等和子集.md | 9 +++++++-- 1 file changed, 7 insertions(+), 2 deletions(-) diff --git a/problems/0416.分割等和子集.md b/problems/0416.分割等和子集.md index 9c90ea27..fa677da0 100644 --- a/problems/0416.分割等和子集.md +++ b/problems/0416.分割等和子集.md @@ -143,12 +143,17 @@ dp[j]的数值一定是小于等于j的。 class Solution { public: bool canPartition(vector& nums) { - // 使用标准库函数 便捷求和 - int sum = accumulate(nums.begin(), nums.end(), 0); + int sum = 0; + // dp[i]中的i表示背包内总和 // 题目中说：每个数组中的元素不会超过 100，数组的大小不会超过 200 // 总和不会大于20000，背包最大只需要其中一半，所以10001大小就可以了 vector dp(10001, 0); + for (int i = 0; i < nums.size(); i++) { + sum += nums[i]; + } + // 也可以使用库函数一步求和 + // int sum = accumulate(nums.begin(), nums.end(), 0); if (sum % 2 == 1) return false; int target = sum / 2;