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[pull] master from youngyangyang04:master #74

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Merged
pull merged 9 commits into AlgorithmAndLeetCode:master from youngyangyang04:master
Sep 1, 2022
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51 changes: 51 additions & 0 deletions problems/0127.单词接龙.md
View file Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -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
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59 changes: 45 additions & 14 deletions problems/0347.前K个高频元素.md
View file Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -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<Integer,Integer> 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<Integer,Integer> 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<Map.Entry<Integer, Integer>> 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<Map.Entry<Integer, Integer>> queue = new PriorityQueue<>((o1, o2) -> o2.getValue() - o1.getValue());
for (Map.Entry<Integer, Integer> 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<int[]> pq = new PriorityQueue<>((pair1, pair2)->pair2[1]-pair1[1]);
for(Map.Entry<Integer,Integer> 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<k;i++){//依次从队头弹出k个,就是出现频率前k高的元素
ans[i] = pq.poll()[0];
}
return result;
return ans;
}
//解法2:基于小顶堆实现
public int[] topKFrequent2(int[] nums, int k) {
Map<Integer,Integer> 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<int[]> pq = new PriorityQueue<>((pair1,pair2)->pair1[1]-pair2[1]);
for(Map.Entry<Integer,Integer> entry:map.entrySet()){//小顶堆只需要维持k个元素有序
if(pq.size()<k){//小顶堆元素个数小于k个时直接加
pq.add(new int[]{entry.getKey(),entry.getValue()});
}else{
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)
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;
}
}
```
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2 changes: 2 additions & 0 deletions problems/0416.分割等和子集.md
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Original file line number Diff line number Diff line change
Expand Up @@ -152,6 +152,8 @@ public:
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;

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