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Merge pull request #1776 from juguagua/leetcode-modify-the-code-of-the-hash

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修改完善哈希表部分代码和注释
This commit is contained in:
程序员Carl
2022-11-25 15:27:26 +08:00
committed by GitHub
parent c434e343a8 1afff79669
commit 35fbcb328a
7 changed files with 88 additions and 103 deletions
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14
problems/0001.两数之和.md
View File

@ -134,12 +134,13 @@ public int[] twoSum(int[] nums, int target) {
}
Map<Integer, Integer> map = new HashMap<>();
for(int i = 0; i < nums.length; i++){
int temp = target - nums[i];
int temp = target - nums[i]; // 遍历当前元素，并在map中寻找是否有匹配的key
if(map.containsKey(temp)){
res[1] = i;
res[0] = map.get(temp);
break;
}
map.put(nums[i], i);
map.put(nums[i], i); // 如果没找到匹配对，就把访问过的元素和下标加入到map中
}
return res;
}
@ -153,15 +154,16 @@ class Solution:
records = dict()
for index, value in enumerate(nums):
if target - value in records:
if target - value in records: # 遍历当前元素，并在map中寻找是否有匹配的key
return [records[target- value], index]
records[value] = index
records[value] = index # 遍历当前元素，并在map中寻找是否有匹配的key
return []
```
Go：
```go
// 暴力解法
func twoSum(nums []int, target int) []int {
for k1, _ := range nums {
for k2 := k1 + 1; k2 < len(nums); k2++ {
@ -216,11 +218,11 @@ Javascript
```javascript
var twoSum = function (nums, target) {
let hash = {};
for (let i = 0; i < nums.length; i++) {
for (let i = 0; i < nums.length; i++) { // 遍历当前元素，并在map中寻找是否有匹配的key
if (hash[target - nums[i]] !== undefined) {
return [i, hash[target - nums[i]]];
}
hash[nums[i]] = i;
hash[nums[i]] = i; // 如果没找到匹配对，就把访问过的元素和下标加入到map中
}
return [];
};

22
problems/0015.三数之和.md
View File

@ -253,13 +253,15 @@ class Solution {
public List<List<Integer>> threeSum(int[] nums) {
List<List<Integer>> result = new ArrayList<>();
Arrays.sort(nums);
// 找出a + b + c = 0
// a = nums[i], b = nums[left], c = nums[right]
for (int i = 0; i < nums.length; i++) {
// 排序之后如果第一个元素已经大于零，那么无论如何组合都不可能凑成三元组，直接返回结果就可以了
if (nums[i] > 0) {
return result;
}
if (i > 0 && nums[i] == nums[i - 1]) {
if (i > 0 && nums[i] == nums[i - 1]) { // 去重a
continue;
}
@ -273,7 +275,7 @@ class Solution {
left++;
} else {
result.add(Arrays.asList(nums[i], nums[left], nums[right]));
// 去重逻辑应该放在找到一个三元组之后，对b 和 c去重
while (right > left && nums[right] == nums[right - 1]) right--;
while (right > left && nums[left] == nums[left + 1]) left++;
@ -294,12 +296,15 @@ class Solution:
ans = []
n = len(nums)
nums.sort()
# 找出a + b + c = 0
# a = nums[i], b = nums[left], c = nums[right]
for i in range(n):
left = i + 1
right = n - 1
# 排序之后如果第一个元素已经大于零，那么无论如何组合都不可能凑成三元组，直接返回结果就可以了
if nums[i] > 0:
break
if i >= 1 and nums[i] == nums[i - 1]:
if i >= 1 and nums[i] == nums[i - 1]: # 去重a
continue
while left < right:
total = nums[i] + nums[left] + nums[right]
@ -309,13 +314,14 @@ class Solution:
left += 1
else:
ans.append([nums[i], nums[left], nums[right]])
# 去重逻辑应该放在找到一个三元组之后，对b 和 c去重
while left != right and nums[left] == nums[left + 1]: left += 1
while left != right and nums[right] == nums[right - 1]: right -= 1
left += 1
right -= 1
return ans
```
Python (v2):
Python (v3):
```python
class Solution:
@ -347,12 +353,15 @@ Go：
func threeSum(nums []int) [][]int {
sort.Ints(nums)
res := [][]int{}
// 找出a + b + c = 0
// a = nums[i], b = nums[left], c = nums[right]
for i := 0; i < len(nums)-2; i++ {
// 排序之后如果第一个元素已经大于零，那么无论如何组合都不可能凑成三元组，直接返回结果就可以了
n1 := nums[i]
if n1 > 0 {
break
}
// 去重a
if i > 0 && n1 == nums[i-1] {
continue
}
@ -361,6 +370,7 @@ func threeSum(nums []int)[][]int{
n2, n3 := nums[l], nums[r]
if n1+n2+n3 == 0 {
res = append(res, []int{n1, n2, n3})
// 去重逻辑应该放在找到一个三元组之后，对b 和 c去重
for l < r && nums[l] == n2 {
l++
}

18
problems/0018.四数之和.md
View File

@ -142,19 +142,20 @@ class Solution {
return result;
}
if (i > 0 && nums[i - 1] == nums[i]) {
if (i > 0 && nums[i - 1] == nums[i]) { // 对nums[i]去重
continue;
}
for (int j = i + 1; j < nums.length; j++) {
if (j > i + 1 && nums[j - 1] == nums[j]) {
if (j > i + 1 && nums[j - 1] == nums[j]) { // 对nums[j]去重
continue;
}
int left = j + 1;
int right = nums.length - 1;
while (right > left) {
// nums[k] + nums[i] + nums[left] + nums[right] > target int会溢出
long sum = (long) nums[i] + nums[j] + nums[left] + nums[right];
if (sum > target) {
right--;
@ -162,7 +163,7 @@ class Solution {
left++;
} else {
result.add(Arrays.asList(nums[i], nums[j], nums[left], nums[right]));
// 对nums[left]和nums[right]去重
while (right > left && nums[right] == nums[right - 1]) right--;
while (right > left && nums[left] == nums[left + 1]) left++;
@ -187,9 +188,9 @@ class Solution:
n = len(nums)
res = []
for i in range(n):
if i > 0 and nums[i] == nums[i - 1]: continue
if i > 0 and nums[i] == nums[i - 1]: continue # 对nums[i]去重
for k in range(i+1, n):
if k > i + 1 and nums[k] == nums[k-1]: continue
if k > i + 1 and nums[k] == nums[k-1]: continue # 对nums[k]去重
p = k + 1
q = n - 1
@ -198,6 +199,7 @@ class Solution:
elif nums[i] + nums[k] + nums[p] + nums[q] < target: p += 1
else:
res.append([nums[i], nums[k], nums[p], nums[q]])
# 对nums[p]和nums[q]去重
while p < q and nums[p] == nums[p + 1]: p += 1
while p < q and nums[q] == nums[q - 1]: q -= 1
p += 1
@ -258,12 +260,12 @@ func fourSum(nums []int, target int) [][]int {
// if n1 > target { // 不能这样写,因为可能是负数
// break
// }
if i > 0 && n1 == nums[i-1] {
if i > 0 && n1 == nums[i-1] { // 对nums[i]去重
continue
}
for j := i + 1; j < len(nums)-2; j++ {
n2 := nums[j]
if j > i+1 && n2 == nums[j-1] {
if j > i+1 && n2 == nums[j-1] { // 对nums[j]去重
continue
}
l := j + 1
@ -320,6 +322,8 @@ var fourSum = function(nums, target) {
if(sum < target) { l++; continue}
if(sum > target) { r--; continue}
res.push([nums[i], nums[j], nums[l], nums[r]]);
// 对nums[left]和nums[right]去重
while(l < r && nums[l] === nums[++l]);
while(l < r && nums[r] === nums[--r]);
}

35
problems/0242.有效的字母异位词.md
View File

@ -101,7 +101,7 @@ class Solution {
int[] record = new int[26];
for (int i = 0; i < s.length(); i++) {
record[s.charAt(i) - 'a']++;
record[s.charAt(i) - 'a']++; // 并不需要记住字符a的ASCII，只要求出一个相对数值就可以了
}
for (int i = 0; i < t.length(); i++) {
@ -109,11 +109,11 @@ class Solution {
}
for (int count: record) {
if (count != 0) {
if (count != 0) { // record数组如果有的元素不为零0，说明字符串s和t 一定是谁多了字符或者谁少了字符。
return false;
}
}
return true;
return true; // record数组所有元素都为零0，说明字符串s和t是字母异位词
}
}
```
@ -166,35 +166,10 @@ class Solution(object):
Go：
```go
func isAnagram(s string, t string) bool {
if len(s)!=len(t){
return false
}
exists := make(map[byte]int)
for i:=0;i<len(s);i++{
if v,ok:=exists[s[i]];v>=0&&ok{
exists[s[i]]=v+1
}else{
exists[s[i]]=1
}
}
for i:=0;i<len(t);i++{
if v,ok:=exists[t[i]];v>=1&&ok{
exists[t[i]]=v-1
}else{
return false
}
}
return true
}
```
Go写法二（没有使用slice作为哈希表，用数组来代替）：
```go
func isAnagram(s string, t string) bool {
record := [26]int{}
for _, r := range s {
record[r-rune('a')]++
}
@ -202,7 +177,7 @@ func isAnagram(s string, t string) bool {
record[r-rune('a')]--
}
return record == [26]int{}
return record == [26]int{} // record数组如果有的元素不为零0，说明字符串s和t 一定是谁多了字符或者谁少了字符。
}
```

32
problems/0349.两个数组的交集.md
View File

@ -147,32 +147,24 @@ Python3：
```python
class Solution:
def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
return list(set(nums1) & set(nums2)) # 两个数组先变成集合，求交集后还原为数组
val_dict = {}
ans = []
for num in nums1:
val_dict[num] = 1
for num in nums2:
if num in val_dict.keys() and val_dict[num] == 1:
ans.append(num)
val_dict[num] = 0
return ans
```
Go：
```go
func intersection(nums1 []int, nums2 []int) []int {
m := make(map[int]int)
for _, v := range nums1 {
m[v] = 1
}
var res []int
// 利用count>0，实现重复值只拿一次放入返回结果中
for _, v := range nums2 {
if count, ok := m[v]; ok && count > 0 {
res = append(res, v)
m[v]--
}
}
return res
}
```
```golang
//优化版，利用set，减少count统计
func intersection(nums1 []int, nums2 []int) []int {
set:=make(map[int]struct{},0)
set:=make(map[int]struct{},0) // 用map模拟set
res:=make([]int,0)
for _,v:=range nums1{
if _,ok:=set[v];!ok{

18
problems/0383.赎金信.md
View File

@ -147,11 +147,11 @@ class Solution:
arr = [0] * 26
for x in magazine:
for x in magazine: # 记录 magazine里各个字符出现次数
arr[ord(x) - ord('a')] += 1
for x in ransomNote:
if arr[ord(x) - ord('a')] == 0:
for x in ransomNote: # 在arr里对应的字符个数做--操作
if arr[ord(x) - ord('a')] == 0: # 如果没有出现过直接返回
return False
else:
arr[ord(x) - ord('a')] -= 1
@ -234,12 +234,12 @@ Go：
```go
func canConstruct(ransomNote string, magazine string) bool {
record := make([]int, 26)
for _, v := range magazine {
for _, v := range magazine { // 通过recode数据记录 magazine里各个字符出现次数
record[v-'a']++
}
for _, v := range ransomNote {
for _, v := range ransomNote { // 遍历ransomNote，在record里对应的字符个数做--操作
record[v-'a']--
if record[v-'a'] < 0 {
if record[v-'a'] < 0 { // 如果小于零说明ransomNote里出现的字符，magazine没有
return false
}
}
@ -258,12 +258,12 @@ javaScript:
var canConstruct = function(ransomNote, magazine) {
const strArr = new Array(26).fill(0),
base = "a".charCodeAt();
for(const s of magazine) {
for(const s of magazine) { // 记录 magazine里各个字符出现次数
strArr[s.charCodeAt() - base]++;
}
for(const s of ransomNote) {
for(const s of ransomNote) { // 对应的字符个数做--操作
const index = s.charCodeAt() - base;
if(!strArr[index]) return false;
if(!strArr[index]) return false; // 如果没记录过直接返回false
strArr[index]--;
}
return true;

8
problems/0454.四数相加II.md
View File

@ -168,13 +168,15 @@ Go：
```go
func fourSumCount(nums1 []int, nums2 []int, nums3 []int, nums4 []int) int {
m := make(map[int]int)
m := make(map[int]int) //key:a+b的数值，value:a+b数值出现的次数
count := 0
// 遍历nums1和nums2数组，统计两个数组元素之和，和出现的次数，放到map中
for _, v1 := range nums1 {
for _, v2 := range nums2 {
m[v1+v2]++
}
}
// 遍历nums3和nums4数组，找到如果 0-(c+d) 在map中出现过的话，就把map中key对应的value也就是出现次数统计出来
for _, v3 := range nums3 {
for _, v4 := range nums4 {
count += m[-v3-v4]
@ -197,14 +199,14 @@ javaScript:
var fourSumCount = function(nums1, nums2, nums3, nums4) {
const twoSumMap = new Map();
let count = 0;
// 统计nums1和nums2数组元素之和，和出现的次数，放到map中
for(const n1 of nums1) {
for(const n2 of nums2) {
const sum = n1 + n2;
twoSumMap.set(sum, (twoSumMap.get(sum) || 0) + 1)
}
}
// 找到如果 0-(c+d) 在map中出现过的话，就把map中key对应的value也就是出现次数统计出来
for(const n3 of nums3) {
for(const n4 of nums4) {
const sum = n3 + n4;