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Commit d14b98b

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Merge branch 'youngyangyang04:master' into zhicheng-lee-patch-3
2 parents 814ea3e + c1cb69f commit d14b98b

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3 files changed

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-14
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3 files changed

+98
-14
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‎problems/0127.单词接龙.md

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Original file line numberDiff line numberDiff line change
@@ -158,6 +158,57 @@ class Solution:
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return 0
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```
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## Go
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```go
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func ladderLength(beginWord string, endWord string, wordList []string) int {
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wordMap, que, depth := getWordMap(wordList, beginWord), []string{beginWord}, 0
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for len(que) > 0 {
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depth++
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qLen := len(que) // 单词的长度
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for i := 0; i < qLen; i++ {
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word := que[0]
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que = que[1:] // 首位单词出队
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candidates := getCandidates(word)
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for _, candidate := range candidates {
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if _, exist := wordMap[candidate]; exist { // 用生成的结果集去查询
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if candidate == endWord {
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return depth + 1
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}
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delete(wordMap, candidate) // 删除集合中的用过的结果
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que = append(que, candidate)
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}
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}
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}
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}
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return 0
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}
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// 获取单词Map为后续的查询增加速度
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func getWordMap(wordList []string, beginWord string) map[string]int {
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wordMap := make(map[string]int)
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for i, word := range wordList {
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if _, exist := wordMap[word]; !exist {
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if word != beginWord {
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wordMap[word] = i
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}
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}
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}
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return wordMap
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}
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// 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
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func getCandidates(word string) []string {
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var res []string
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for i := 0; i < 26; i++ {
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for j := 0; j < len(word); j++ {
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if word[j] != byte(int('a')+i) {
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res = append(res, word[:j]+string(int('a')+i)+word[j+1:])
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}
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}
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}
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return res
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}
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```
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## JavaScript
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```javascript

‎problems/0347.前K个高频元素.md

Lines changed: 45 additions & 14 deletions
Original file line numberDiff line numberDiff line change
@@ -140,24 +140,55 @@ public:
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Java:
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```java
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/*Comparator接口说明:
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* 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
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* 对于队列:排在前面意味着往队头靠
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* 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),
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* 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
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* */
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class Solution {
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public int[] topKFrequent(int[] nums, int k) {
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int[] result = new int[k];
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HashMap<Integer,Integer> map = new HashMap<>();
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for(int num : nums){
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map.put(num,map.getOrDefault(num, 0) + 1);
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//解法1:基于大顶堆实现
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public int[] topKFrequent1(int[] nums, int k) {
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Map<Integer,Integer> map = new HashMap<>();//key为数组元素值,val为对应出现次数
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for(int num:nums){
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map.put(num,map.getOrDefault(num,0)+1);
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}
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Set<Map.Entry<Integer, Integer>> entries = map.entrySet();
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// 根据map的value值,构建于一个大顶堆(o1 - o2: 小顶堆, o2 - o1 : 大顶堆)
153-
PriorityQueue<Map.Entry<Integer, Integer>> queue = new PriorityQueue<>((o1, o2) -> o2.getValue() - o1.getValue());
154-
for (Map.Entry<Integer, Integer> entry : entries) {
155-
queue.offer(entry);
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//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
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//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)
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PriorityQueue<int[]> pq = new PriorityQueue<>((pair1, pair2)->pair2[1]-pair1[1]);
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for(Map.Entry<Integer,Integer> entry:map.entrySet()){//大顶堆需要对所有元素进行排序
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pq.add(new int[]{entry.getKey(),entry.getValue()});
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}
157-
for (int i = k - 1; i >= 0; i--) {
158-
result[i] = queue.poll().getKey();
162+
int[] ans = new int[k];
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for(int i=0;i<k;i++){//依次从队头弹出k个,就是出现频率前k高的元素
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ans[i] = pq.poll()[0];
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}
160-
return result;
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return ans;
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}
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//解法2:基于小顶堆实现
169+
public int[] topKFrequent2(int[] nums, int k) {
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Map<Integer,Integer> map = new HashMap<>();//key为数组元素值,val为对应出现次数
171+
for(int num:nums){
172+
map.put(num,map.getOrDefault(num,0)+1);
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}
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//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
175+
//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)
176+
PriorityQueue<int[]> pq = new PriorityQueue<>((pair1,pair2)->pair1[1]-pair2[1]);
177+
for(Map.Entry<Integer,Integer> entry:map.entrySet()){//小顶堆只需要维持k个元素有序
178+
if(pq.size()<k){//小顶堆元素个数小于k个时直接加
179+
pq.add(new int[]{entry.getKey(),entry.getValue()});
180+
}else{
181+
if(entry.getValue()>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)
182+
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
183+
pq.add(new int[]{entry.getKey(),entry.getValue()});
184+
}
185+
}
186+
}
187+
int[] ans = new int[k];
188+
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
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ans[i] = pq.poll()[0];
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}
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return ans;
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}
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}
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```

‎problems/0416.分割等和子集.md

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for (int i = 0; i < nums.size(); i++) {
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sum += nums[i];
154154
}
155+
// 也可以使用库函数一步求和
156+
// int sum = accumulate(nums.begin(), nums.end(), 0);
155157
if (sum % 2 == 1) return false;
156158
int target = sum / 2;
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