JavaScript 面试备忘单 - 第 2 部分

来源:undefined 2025-02-03 01:50:39 1042

常见 leetcode 模式

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

37

38

39

40

41

42

43

44

45

46

47

48

49

50

51

52

53

54

55

56

57

58

59

60

61

62

63

64

65

66

67

68

69

70

71

72

73

74

75

76

77

78

79

80

81

82

83

84

85

86

87

88

89

90

91

92

93

94

95

96

97

98

99

100

101

102

103

104

105

106

107

108

109

110

111

112

113

114

115

116

117

118

119

120

121

122

123

124

125

126

127

128

129

130

131

132

133

134

135

136

137

138

139

140

141

142

143

144

145

146

147

148

149

150

151

152

153

154

155

156

157

158

159

160

161

162

163

164

165

166

167

168

169

170

171

172

173

174

175

176

177

178

179

180

181

182

183

184

185

186

187

188

189

190

191

192

193

194

195

196

197

198

199

200

201

202

203

204

205

206

207

208

209

210

211

212

213

214

215

216

217

218

219

220

221

222

223

224

225

226

227

228

229

230

231

232

233

234

235

236

237

238

239

240

241

242

243

244

245

246

247

248

249

250

251

252

253

254

255

256

257

258

259

260

261

262

263

264

265

266

267

268

269

270

271

272

273

274

// two pointers - in-place array modification

const modifyarray = (arr) => {

let writepointer = 0;

for (let readpointer = 0; readpointer < arr.length; readpointer++) {

if (/* condition */) {

[arr[writepointer], arr[readpointer]] = [arr[readpointer], arr[writepointer]];

writepointer++;

}

}

return writepointer; // often returns new length or modified position

};

// fast and slow pointers (floyds cycle detection)

const hascycle = (head) => {

let slow = head, fast = head;

while (fast && fast.next) {

slow = slow.next;

fast = fast.next.next;

if (slow === fast) return true;

}

return false;

};

// sliding window - fixed size

const fixedslidingwindow = (arr, k) => {

let sum = 0;

// initialize first window

for (let i = 0; i < k; i++) {

sum += arr[i];

}

let maxsum = sum;

// slide window

for (let i = k; i < arr.length; i++) {

sum = sum - arr[i - k] + arr[i];

maxsum = math.max(maxsum, sum);

}

return maxsum;

};

// sliding window - variable size

const varslidingwindow = (arr, target) => {

let start = 0, sum = 0, minlen = infinity;

for (let end = 0; end < arr.length; end++) {

sum += arr[end];

while (sum >= target) {

minlen = math.min(minlen, end - start + 1);

sum -= arr[start];

start++;

}

}

return minlen === infinity ? 0 : minlen;

};

// bfs - level order traversal

const levelorder = (root) => {

if (!root) return [];

const result = [];

const queue = [root];

while (queue.length) {

const levelsize = queue.length;

const currentlevel = [];

for (let i = 0; i < levelsize; i++) {

const node = queue.shift();

currentlevel.push(node.val);

if (node.left) queue.push(node.left);

if (node.right) queue.push(node.right);

}

result.push(currentlevel);

}

return result;

};

// dfs - recursive template

const dfs = (root) => {

const result = [];

const traverse = (node) => {

if (!node) return;

// pre-order

result.push(node.val);

traverse(node.left);

// in-order would be here

traverse(node.right);

// post-order would be here

};

traverse(root);

return result;

};

// backtracking template

const backtrack = (nums) => {

const result = [];

const bt = (path, choices) => {

if (/* ending condition */) {

result.push([...path]);

return;

}

for (let i = 0; i < choices.length; i++) {

// make choice

path.push(choices[i]);

// recurse

bt(path, /* remaining choices */);

// undo choice

path.pop();

}

};

bt([], nums);

return result;

};

// dynamic programming - bottom up template

const dpbottomup = (n) => {

const dp = new array(n + 1).fill(0);

dp[0] = 1; // base case

for (let i = 1; i <= n; i++) {

for (let j = 0; j < i; j++) {

dp[i] += dp[j] * /* some calculation */;

}

}

return dp[n];

};

// dynamic programming - top down template

const dptopdown = (n) => {

const memo = new map();

const dp = (n) => {

if (n <= 1) return 1;

if (memo.has(n)) return memo.get(n);

let result = 0;

for (let i = 0; i < n; i++) {

result += dp(i) * /* some calculation */;

}

memo.set(n, result);

return result;

};

return dp(n);

};

// monotonic stack template

const monotonicstack = (arr) => {

const stack = []; // [index, value]

const result = new array(arr.length).fill(-1);

for (let i = 0; i < arr.length; i++) {

while (stack.length && stack[stack.length - 1][1] > arr[i]) {

const [previndex, _] = stack.pop();

result[previndex] = i - previndex;

}

stack.push([i, arr[i]]);

}

return result;

};

// prefix sum

const prefixsum = (arr) => {

const prefix = [0];

for (let i = 0; i < arr.length; i++) {

prefix.push(prefix[prefix.length - 1] + arr[i]);

}

// sum of range [i, j] = prefix[j + 1] - prefix[i]

return prefix;

};

// binary search variations

const binarysearchleftmost = (arr, target) => {

let left = 0, right = arr.length;

while (left < right) {

const mid = math.floor((left + right) / 2);

if (arr[mid] < target) left = mid + 1;

else right = mid;

}

return left;

};

const binarysearchrightmost = (arr, target) => {

let left = 0, right = arr.length;

while (left < right) {

const mid = math.floor((left + right) / 2);

if (arr[mid] <= target) left = mid + 1;

else right = mid;

}

return left - 1;

};

// trie operations

class trienode {

constructor() {

this.children = new map();

this.isendofword = false;

}

}

class trie {

constructor() {

this.root = new trienode();

}

insert(word) {

let node = this.root;

for (const char of word) {

if (!node.children.has(char)) {

node.children.set(char, new trienode());

}

node = node.children.get(char);

}

node.isendofword = true;

}

search(word) {

let node = this.root;

for (const char of word) {

if (!node.children.has(char)) return false;

node = node.children.get(char);

}

return node.isendofword;

}

startswith(prefix) {

let node = this.root;

for (const char of prefix) {

if (!node.children.has(char)) return false;

node = node.children.get(char);

}

return true;

}

}

// union find (disjoint set)

class unionfind {

constructor(n) {

this.parent = array.from({length: n}, (_, i) => i);

this.rank = new array(n).fill(0);

}

find(x) {

if (this.parent[x] !== x) {

this.parent[x] = this.find(this.parent[x]); // path compression

}

return this.parent[x];

}

union(x, y) {

let rootx = this.find(x);

let rooty = this.find(y);

if (rootx !== rooty) {

if (this.rank[rootx] < this.rank[rooty]) {

[rootx, rooty] = [rooty, rootx];

}

this.parent[rooty] = rootx;

if (this.rank[rootx] === this.rank[rooty]) {

this.rank[rootx]++;

}

}

}

}

登录后复制

常见的时间/空间复杂度模式

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

// O(1) - Constant

Array.push(), Array.pop(), Map.set(), Map.get()

// O(log n) - Logarithmic

Binary Search, Balanced BST operations

// O(n) - Linear

Array traversal, Linear Search

// O(n log n) - Linearithmic

Efficient sorting (Array.sort())

// O(n²) - Quadratic

Nested loops, Simple sorting algorithms

// O(2ⁿ) - Exponential

Recursive solutions without memoization

登录后复制

以上就是JavaScript 面试备忘单 - 第 2 部分的详细内容,更多请关注php中文网其它相关文章!

最新文章