|
1 | | -学习笔记 |
| 1 | +#### XOR异或 ^ |
| 2 | +``` |
| 3 | +x ^ 0 = x |
| 4 | +# 1s = ~0 |
| 5 | +x ^ 1s = ~x |
| 6 | +x ^ (~x) = 1s |
| 7 | +x ^ x = 0 |
| 8 | +c = a ^ b --> a ^ c = b --> b ^ c = a |
| 9 | +a ^ b ^ c = a ^ (b ^ c) = (a ^ b) ^ c |
| 10 | +``` |
| 11 | +<br/> |
| 12 | + |
| 13 | +### 指定位置的位运算 |
| 14 | +``` |
| 15 | +# 将x最右边的n位清零 |
| 16 | +x & (~0 << n) |
| 17 | +# 获取x第n位的值(0 or 1) |
| 18 | +x >> n & 1 |
| 19 | +# 获取第n位的幂值 |
| 20 | +x & (1 << n) |
| 21 | +# 仅将第n位置为 1 |
| 22 | +x | (1 << n) |
| 23 | +# 仅将第n位置为 0 |
| 24 | +x & (~(1 << n) |
| 25 | +# 将x最高位至n位(含n)清零 |
| 26 | +x & ((1 << n) - 1) |
| 27 | +# 将x第n位至0位(含n)清零 |
| 28 | +x & (~((1 << n + 1) - 1)) |
| 29 | +``` |
| 30 | +<br/> |
| 31 | + |
| 32 | +### 实战运用 |
| 33 | +``` |
| 34 | +# 清零最低位 1 |
| 35 | +X = X & (X - 1) |
| 36 | +# 得到最低位 1 |
| 37 | +X = X & -X |
| 38 | +``` |
| 39 | +<br/> |
| 40 | + |
| 41 | +### 布隆过滤器 |
| 42 | +``` |
| 43 | +# 核心: |
| 44 | + 超大位数组 + hash函数 |
| 45 | +
|
| 46 | +# 添加元素 |
| 47 | + 1. 将添加元素给k个hash函数 |
| 48 | + 2. 得到对应位数组的k个位置 |
| 49 | + 3. 将对应位置设为 1 |
| 50 | +# 查询元素 |
| 51 | + 1. 将下旬元素给k个hash函数 |
| 52 | + 2. 得到对应位数组的k个位置 |
| 53 | + 3. 只要有一个位置值为 0, 则元素不存在 |
| 54 | + 3. 如果k个位置值全为 1, 则元素可能存在(存在误判, 用于最外层过滤) |
| 55 | +``` |
| 56 | + |
| 57 | +``` |
| 58 | +/** |
| 59 | + * 示例代码 |
| 60 | + */ |
| 61 | +public class BloomFilter { |
| 62 | + private static final int DEFAULT_SIZE = 2 << 24; |
| 63 | + private static final int[] seeds = new int[] { 5, 7, 11, 13, 31, 37, 61 }; |
| 64 | + private BitSet bits = new BitSet(DEFAULT_SIZE); |
| 65 | + private SimpleHash[] func = new SimpleHash[seeds.length]; |
| 66 | + public BloomFilter() { |
| 67 | + for (int i = 0; i < seeds.length; i++) { |
| 68 | + func[i] = new SimpleHash(DEFAULT_SIZE, seeds[i]); |
| 69 | + } |
| 70 | + } |
| 71 | + public void add(String value) { |
| 72 | + for (SimpleHash f : func) { |
| 73 | + bits.set(f.hash(value), true); |
| 74 | + } |
| 75 | + } |
| 76 | + public boolean contains(String value) { |
| 77 | + if (value == null) { |
| 78 | + return false; |
| 79 | + } |
| 80 | + boolean ret = true; |
| 81 | + for (SimpleHash f : func) { |
| 82 | + ret = ret && bits.get(f.hash(value)); |
| 83 | + } |
| 84 | + return ret; |
| 85 | + } |
| 86 | + // 内部类,simpleHash |
| 87 | + public static class SimpleHash { |
| 88 | + private int cap; |
| 89 | + private int seed; |
| 90 | + public SimpleHash(int cap, int seed) { |
| 91 | + this.cap = cap; |
| 92 | + this.seed = seed; |
| 93 | + } |
| 94 | + public int hash(String value) { |
| 95 | + int result = 0; |
| 96 | + int len = value.length(); |
| 97 | + for (int i = 0; i < len; i++) { |
| 98 | + result = seed * result + value.charAt(i); |
| 99 | + } |
| 100 | + return (cap - 1) & result; |
| 101 | + } |
| 102 | + } |
| 103 | +} |
| 104 | +``` |
| 105 | +<br/> |
| 106 | + |
| 107 | +### LRU Cache |
| 108 | +```java |
| 109 | +/** |
| 110 | + * 示例代码 |
| 111 | + */ |
| 112 | +class LRUCache { |
| 113 | + /** |
| 114 | + * 缓存映射表 |
| 115 | + */ |
| 116 | + private Map<Integer, DLinkNode> cache = new HashMap<>(); |
| 117 | + /** |
| 118 | + * 缓存大小 |
| 119 | + */ |
| 120 | + private int size; |
| 121 | + /** |
| 122 | + * 缓存容量 |
| 123 | + */ |
| 124 | + private int capacity; |
| 125 | + /** |
| 126 | + * 链表头部和尾部 |
| 127 | + */ |
| 128 | + private DLinkNode head, tail; |
| 129 | + |
| 130 | + public LRUCache(int capacity) { |
| 131 | + //初始化缓存大小,容量和头尾节点 |
| 132 | + this.size = 0; |
| 133 | + this.capacity = capacity; |
| 134 | + head = new DLinkNode(); |
| 135 | + tail = new DLinkNode(); |
| 136 | + head.next = tail; |
| 137 | + tail.prev = head; |
| 138 | + } |
| 139 | + |
| 140 | + /** |
| 141 | + * 获取节点 |
| 142 | + * @param key 节点的键 |
| 143 | + * @return 返回节点的值 |
| 144 | + */ |
| 145 | + public int get(int key) { |
| 146 | + DLinkNode node = cache.get(key); |
| 147 | + if (node == null) { |
| 148 | + return -1; |
| 149 | + } |
| 150 | + //移动到链表头部 |
| 151 | + (node); |
| 152 | + return node.value; |
| 153 | + } |
| 154 | + |
| 155 | + /** |
| 156 | + * 添加节点 |
| 157 | + * @param key 节点的键 |
| 158 | + * @param value 节点的值 |
| 159 | + */ |
| 160 | + public void put(int key, int value) { |
| 161 | + DLinkNode node = cache.get(key); |
| 162 | + if (node == null) { |
| 163 | + DLinkNode newNode = new DLinkNode(key, value); |
| 164 | + cache.put(key, newNode); |
| 165 | + //添加到链表头部 |
| 166 | + addToHead(newNode); |
| 167 | + ++size; |
| 168 | + //如果缓存已满,需要清理尾部节点 |
| 169 | + if (size > capacity) { |
| 170 | + DLinkNode tail = removeTail(); |
| 171 | + cache.remove(tail.key); |
| 172 | + --size; |
| 173 | + } |
| 174 | + } else { |
| 175 | + node.value = value; |
| 176 | + //移动到链表头部 |
| 177 | + moveToHead(node); |
| 178 | + } |
| 179 | + } |
| 180 | + |
| 181 | + /** |
| 182 | + * 删除尾结点 |
| 183 | + * |
| 184 | + * @return 返回删除的节点 |
| 185 | + */ |
| 186 | + private DLinkNode removeTail() { |
| 187 | + DLinkNode node = tail.prev; |
| 188 | + removeNode(node); |
| 189 | + return node; |
| 190 | + } |
| 191 | + |
| 192 | + /** |
| 193 | + * 删除节点 |
| 194 | + * @param node 需要删除的节点 |
| 195 | + */ |
| 196 | + private void removeNode(DLinkNode node) { |
| 197 | + node.next.prev = node.prev; |
| 198 | + node.prev.next = node.next; |
| 199 | + } |
| 200 | + |
| 201 | + /** |
| 202 | + * 把节点添加到链表头部 |
| 203 | + * |
| 204 | + * @param node 要添加的节点 |
| 205 | + */ |
| 206 | + private void addToHead(DLinkNode node) { |
| 207 | + node.prev = head; |
| 208 | + node.next = head.next; |
| 209 | + head.next.prev = node; |
| 210 | + head.next = node; |
| 211 | + } |
| 212 | + |
| 213 | + /** |
| 214 | + * 把节点移动到头部 |
| 215 | + * @param node 需要移动的节点 |
| 216 | + */ |
| 217 | + private void moveToHead(DLinkNode node) { |
| 218 | + removeNode(node); |
| 219 | + addToHead(node); |
| 220 | + } |
| 221 | + |
| 222 | + /** |
| 223 | + * 链表节点类 |
| 224 | + */ |
| 225 | + private static class DLinkNode { |
| 226 | + Integer key; |
| 227 | + Integer value; |
| 228 | + DLinkNode prev; |
| 229 | + DLinkNode next; |
| 230 | + |
| 231 | + DLinkNode() { |
| 232 | + } |
| 233 | + |
| 234 | + DLinkNode(Integer key, Integer value) { |
| 235 | + this.key = key; |
| 236 | + this.value = value; |
| 237 | + } |
| 238 | + } |
| 239 | +} |
| 240 | +``` |
| 241 | +<br/> |
| 242 | + |
| 243 | +### <a href="https://www.cnblogs.com/onepixel/p/7674659.html">排序算法</a> |
| 244 | + |
| 245 | +#### 冒泡排序 |
| 246 | +```java |
| 247 | +public void bubbleSort(int[] nums) { |
| 248 | + int len = nums.length; |
| 249 | + // 相邻两个数比较, 顺序错误则交换位置 |
| 250 | + // time:O(n ^ 2) space:O(1) |
| 251 | + for (int i = 0; i < len; i++) { |
| 252 | + for (int j = 0; j < len - 1 - i; j++) { |
| 253 | + if (nums[j] > nums[j + 1]) { |
| 254 | + int temp = nums[j]; |
| 255 | + nums[j] = nums[j + 1]; |
| 256 | + nums[j + 1] = temp; |
| 257 | + } |
| 258 | + } |
| 259 | + } |
| 260 | +} |
| 261 | +``` |
| 262 | +<br/> |
| 263 | + |
| 264 | +#### 选择排序 |
| 265 | +```java |
| 266 | +public void selectionSort(int[] nums) { |
| 267 | + int len = nums.length; |
| 268 | + // 在未排序序列中找到最小(大)元素,存放到排序序列的起始位置 |
| 269 | + // time:O(n ^ 2) space:O(1) |
| 270 | + for (int i = 0; i < len; i++) { |
| 271 | + int minIndex = i; |
| 272 | + for (int j = i + 1; j < len; j++) { |
| 273 | + // 找从i开始后的最小数的索引 |
| 274 | + if (nums[j] < nums[minIndex]) minIndex = j; |
| 275 | + } |
| 276 | + // 交换位置 |
| 277 | + int temp = nums[i]; |
| 278 | + nums[i] = nums[minIndex]; |
| 279 | + nums[minIndex] = temp; |
| 280 | + } |
| 281 | +} |
| 282 | +``` |
| 283 | +<br/> |
| 284 | + |
| 285 | +#### 插入排序 |
| 286 | +```java |
| 287 | +public void selectionSort(int[] nums) { |
| 288 | + int len = nums.length; |
| 289 | + // 构建有序序列, 对于未排序数据, 在已排序序列中从后向前扫描, 找到相应位置并插入 |
| 290 | + // time:O(n ^ 2) space:O(1) |
| 291 | + for (int i = 0; i < len; i++) { |
| 292 | + int curNum = nums[i], curIndex = i - 1; |
| 293 | + while (curIndex >= 0 && nums[curIndex] > curNum) { |
| 294 | + nums[curIndex + 1] = nums[curIndex]; |
| 295 | + curIndex--; |
| 296 | + } |
| 297 | + nums[curIndex + 1] = curNum; |
| 298 | + } |
| 299 | +} |
| 300 | +``` |
| 301 | +<br/> |
| 302 | + |
| 303 | +#### 快速排序 |
| 304 | +```java |
| 305 | +/** |
| 306 | + * 通过一趟排序将待排记录分隔成独立的两部分, |
| 307 | + * 其中一部分记录的关键字均比另一部分的关键字小, |
| 308 | + * 则可分别对这两部分记录继续进行排序,以达到整个序列有序 |
| 309 | + */ |
| 310 | +public void quickSort(int[] nums) { |
| 311 | + // time:O(nlog n) space:O(nlog n) |
| 312 | + quickSort(nums, 0, nums.length - 1); |
| 313 | +} |
| 314 | + |
| 315 | +private void quickSort(int[] nums, int begin, int end) { |
| 316 | + if (begin <= end) return; |
| 317 | + // 寻找标杆位置 |
| 318 | + int pivot = partition(nums, begin, end); |
| 319 | + // pivot左边元素下探 |
| 320 | + quickSort(nums, begin, pivot - 1); |
| 321 | + // pivot右边元素下探 |
| 322 | + quickSort(nums, pivot + 1, end); |
| 323 | +} |
| 324 | + |
| 325 | +private int partition(int[] nums, int begin, int end) { |
| 326 | + // pivot: 标杆位置, counter: 小于pivot的元素的个数 |
| 327 | + int pivot = end, counter = begin; |
| 328 | + for (int i = begin; i < end; i++) { |
| 329 | + if (nums[i] < nums[pivot]) { |
| 330 | + int temp = nums[i]; nums[i] = nums[counter]; nums[counter] = temp; |
| 331 | + counter++; |
| 332 | + } |
| 333 | + } |
| 334 | + int temp = nums[pivot]; nums[pivot] = nums[counter]; nums[counter] = temp; |
| 335 | + return counter; |
| 336 | +} |
| 337 | +``` |
| 338 | +<br/> |
| 339 | + |
| 340 | +#### 归并排序 |
| 341 | +```java |
| 342 | +/** |
| 343 | + * 采用分治法(Divide and Conquer)的一个非常典型的应用。 |
| 344 | + * 将已有序的子序列合并,得到完全有序的序列; |
| 345 | + * 即先使每个子序列有序,再使子序列段间有序 |
| 346 | + */ |
| 347 | +public void mergeSort(int[] nums) { |
| 348 | + // time:O(nlog n) space:O(n) |
| 349 | + mergeSort(nums, 0, nums.length - 1); |
| 350 | +} |
| 351 | + |
| 352 | +private void mergeSort(int[] nums, int left, int right) { |
| 353 | + if (left <= right) return; |
| 354 | + int mid = ((right - left) >> 1) + left; |
| 355 | + // 左半部分下探 |
| 356 | + mergeSort(nums, left, mid); |
| 357 | + // 右半部分下探 |
| 358 | + mergeSort(nums, mid + 1, right); |
| 359 | + merge(nums, left, mid, right); |
| 360 | +} |
| 361 | + |
| 362 | +/** |
| 363 | + * 合并两个有序数组 |
| 364 | + */ |
| 365 | +private int merge(int[] nums, int left, int mid, int right) { |
| 366 | + int[] temp = new int[right - left + 1]; |
| 367 | + int i = left, j = mid + 1, k = 0; |
| 368 | + while (i <= mid && j <= right) { |
| 369 | + temp[k++] = nums[i] < nums[j] ? nums[i++] : nums[j++]; |
| 370 | + } |
| 371 | + while (i <= left) temp[k++] = nums[i++]; |
| 372 | + while (j <= right) temp[k++] = nums[j++]; |
| 373 | + System.arraycopy(temp, 0, nums, left, right - left + 1); |
| 374 | +} |
| 375 | +``` |
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