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怎么制作一个简单的网站,wordpress如何修改密码,适合夫妻二人观看的电视剧,有哪些可以做包装袋的网站贪心算法 1)Greedy algorithm 称之为贪心算法或者贪婪算法#xff0c;核心思想是 将寻找最优解的问题分为若干个步骤每一步骤都采用贪心原则#xff0c;选取当前最优解因为未考虑所有可能#xff0c;局部最优的堆叠不一定得到最终解最优 贪心算法例子 Dijkstra while …贪心算法 1)Greedy algorithm 称之为贪心算法或者贪婪算法核心思想是 将寻找最优解的问题分为若干个步骤每一步骤都采用贪心原则选取当前最优解因为未考虑所有可能局部最优的堆叠不一定得到最终解最优 贪心算法例子 Dijkstra while (!list.isEmpty()) {// 选取当前【距离最小】的顶点Vretex curr chooseMinDistVertex(list);// 更新当前顶点到相邻顶点距离updateNeighboursDish(curr);// list集合中移除当前顶点list.remove(curr);// 标记当前顶点已被访问curr.visited true; }未能找到最短路径存在负边 情况得不到正确解贪心原则会认为本次已找到该顶点的最短路径使得该顶点赋为已访问 与之对比Bellman-Ford算法并未考虑局部距离最小顶点而是每次处理所有边 不会出错当然效率不如Dijkstra算法 prim while (!list.isEmpty()) {// 选取当前【距离最小】的顶点Vretex curr chooseMinDistVertex(list);// 更新当前顶点到相邻顶点距离updateNeighboursDish(curr);// list集合中移除当前顶点list.remove(curr);// 标记当前顶点已被访问curr.visited true; }prim与Dijkstra的区别在于根据距离选取最小顶点不同Dijkstra算法是一个累加距离而prim算法中距离是跟相邻顶点间的距离 。 KrusKal ListString list new ArrayList(); DisjoinSet set new DisjoinSet(size); while (list.size() size - 1) {// 选取当前【距离最短】的边Edge poll queue.poll();int i set.find(poll.start);int j set.find(poll.end);// 判断两个集合是否相交if (i ! j) {// 未相交list.add(poll);set.union(i, j); // 相交操作}}其他贪心算法例子 选择排序、堆排序拓扑排序并查集和中的union by size 和union by height哈夫曼编码钱币找零任务编排近似算法 零钱兑换ⅡLeetcode518 递归实现 public class Leetcode518 {public int change(int[] coins, int amount) {return coinChange(coins, 0, amount, new LinkedList(), true);}public int coinChange(int[] coins, int index, int amount, LinkedListInteger stack, boolean first) {if (!first) {stack.push(coins[i]);}int sum 0;if (amount 0) {System.out.printlin(stack);sum 1;} else if (amount 0) {System.out.println(stack)sum 0;} else {for(int i index; i coins.length; i) {sum coinChange(coins, i, amount - coins[i], stack, false);}}if (!stack.isEmpty()) {stack.pop();}return sum;} } 零钱兑换Leetcode322 递归实现 public class Leetcode322 {static int min -1;public int change(int[] coins, int amount) {coinChange(coins, 0, amount, new AtomicInteger(-1), new LinkedListInteger(), true);return min;}public void coinChange(int[] coins, int index, int amount, AtomicInteger count, LinkedListInteger stack, boolean first) {if (!first) {stack.push(coins[i]);}count.incrementAndGet(); // count;int sum 0;if (amount 0) {System.out.println(stack);if (min -1) {min min.get();} else {min Math.min(min, count.get());}} else {for(int i index; i coins.length; i) {sum coinChange(coins, i, amount - coins[i], count, stack, false);}}count.decrementAndGet(); // count–; if (!stack.isEmpty()) {stack.pop();}return sum;}public static void main(String[] args) {int[] coins {5, 2, 1};Leetcode322 leetcode new Leetcode322();System.out.printlin(leetcode.coinChange(coins, 5));} }贪心实现 public class Leetcode322{public int coinChange(int[] coins, int amount) {// 前提是coins是降序排序int reminder amount;int count;for(int coin : coins) {while (reminder coin) {reminder - coin;count;}if (reminder coin) {reminder - coin;count;break;}}if (reminder 0) {return -1;} else {return count;}}}但是这个代码放到Leetcode上跑有些测试用例是不能通过。 动态规划实现 使用动态规划求解如下面代码 public class Leetcode322{public int coinChange(int[] coins, int amount) {int[] dp new int[amount 1];Arrays.fill(dp, amount 1);dp[0] 0;for(int coin : coins) {for(int j coin; j amount 1; j) {dp[j] Math.min(dp[j], dp[j - coin] 1);}}int r dp[amount];return r amount ? -1 : r;}}哈夫曼编码 Huffman树构建过程 统计出现频次字符放入优先级队列每次出对两个频次最低元素(小顶堆)当队列中只有一个元素时构建Huffman树完成 public class Huffman{static class Node{Character ch;int freq;Node left;Node right;String code;public Node(Character ch) {this.ch ch;}public Node(int freq, Node left, Node right) {this.freq freq;this.left left;this.right right;}int getFreq() {return freq;}boolean isLeaf() {return node.left null;}Overridepublic void toString() {return Node{ ch ch , freq freq };}}String str;Node root;HashMapCharacter, Node map new HashMap();public Huffman(String str) {this.str str;char[] chars str.toCharArray();// 1.统计频次for(char ch : chars) {if (!map.containsKey(ch)) {map.put(ch, new Node(ch));} else {Node node map.get(ch);node.freq;}//方法引用// Node node map.computeIfAbsent(ch, Node :: new);// node.frea;}for(Node node : map.values()) {System.out.println(node.toString());}2.构造树PriorityQueueNode queue new PriorityQueue(Comparator.ComparingInt(Node::getFreq));queue.offerAll(map.values());while (queue.size() 2) {Node x queue.poll();Node y queue.poll();int f x.freq y.freq;queue.offer(new Node(f, x, y));}root queue.poll();System.out.println(root);// 功能3 计算每个字符编码 //功能4 字符串编码后占几个bitint sum dfs(root, new StringBuilder()); // 得到字符串编码后占的bitfor(Node node : map.values()) {System.out.printlin(node);}System.out.println(sum);}public int dfs(Node node, StringBuilder sb) {int sum 0;if (node.isLeaf()) {//编码node.node sb.toString();sum sb.length() * node.freq;// System.out.println(sb.toString()); } else {sum dfs(node.left, sb.append(0));sum dfs(node.right, sb.append(1));}if (sb.length() 0) {sb.deleteCharAt(sb.length() - 1);}return sum;}public String encode() {char[] chars str.toCharArray();StringBuilder sb new StringBuilder();for(char ch : chars) {sb.append(map.get(ch).code);}return sb.toString();}public String decode(String str) {int i 0;char[] chars str.toCharArray();StringBuilder sb new StringBuilder();Node node root;while (i chars.length) {if (!node.isLeaf()) {if (chars[i] 0) {node node.left;} else if (chars[i] 1){node node.right;}i;} if (node.isLeaf()) {sb.append(node.ch);node root;}}return sb.toString();} }活动选择问题 要在一个会议室举办n个活动 每个活动有它们各自的起始和结束时间找出时间上不冲突的活动组合能够最充分利用会议室(举办的活动次数最多) public class ActivitySelectionProblem{static class Activity{int index;int start;int end;public Activity(int index, int start, int end) {this.index index;this.start start;this.end end;}public int getEnd() {return end;}Overridepublic void tostring() {return Activity( index );}}public static void main(String[] args) {Activity[] activity new Activity[]{new Activity(0, 1, 3),new Activity(1, 2, 4),new Activity(2, 3, 5)} Arrays.sort(activity, Comparator.comparingInt(Activity::getEnd))System.out.println(activity);// select(activity, activity.length);}public void select(Activity[] activity, int n) {Listint[] res new ArrayList();res.add(activity[0]);Activity pre res;for(int i 1; i n; i) {Activity curr activity[i];if (curr.start pre.end) {res.add(curr);pre curr;}}for(Activity a : res) {System.out.println(a);}} }Leetcode435 无重叠区间 public class Leetcode435{public int eraseOverlapIntervals(int[][] intervals) {// 根据数组中的第二个元素先升序排序Arrays.sort(intervals, (a, b) - a[1] - b[1]);Listint[] res new ArrayList();res.add(intervals[0]);int[] pre res.get(0);int count 0; // 减少个数for(int i 1; i intervals.length; i) {int[] curr intervals[i];if (curr[0] pre[1]) { // 当前的初始值小于前一个的结束值有冲突count;} else { // 只要当前的初始值大于等于前一个的结束值则不冲突res.add(curr);pre curr;}}return count;} }分数背包问题 n个液体物品有重量和价格属性现取走10L物品每次可以不拿全拿拿一部分问最高价值是多少
public class FractionalKnapsackProblem{static class Item{int index;int weight;int value;public Item(int index, int weight, int value) {this.index index;this.weight weight;this.value value;}public int perValue() {return value / weight;}Overridepublic void toString() {return Item( index );}}public static void main(String[] args) {Item[] items new Item[]{new Item(0, 4, 24),new Item(1, 8, 160),new Item(2, 2, 4000),new Item(3, 6, 108),new Item(4, 1, 4000),}select(items, 10);}public int select(Item[] items, int n) {Arrays.sort(items, Comparator.comparingInt(Item::preValue).reverse());int sum 0;for(int i 0; i items.length; i) {Item curr items[i];if (n curr.weight) {n - curr.weight;sum curr.value;} else {sum curr.perValue() * n;break;}}return sum;} }0-1背包问题 n个物体都是固体有重量和价值现取走不超过10克物品每次可以不拿或者全拿问最高价值是多少 public class FractionalKnapsackProblem{static class Item{int index;int weight;int value;public Item(int index, int weight, int value) {this.index index;this.weight weight;this.value value;}public int perValue() {return value / weight;}Overridepublic void toString() {return Item( index );}}public static void main(String[] args) {Item[] items new Item[]{new Item(0, 1, 1000000),new Item(1, 4, 1600),new Item(2, 8, 2400),new Item(3, 5, 30),}select(items, 10);}public int select(Item[] items, int n) {Arrays.sort(items, Comparator.comparingInt(Item::preValue).reverse());int sum 0;for(int i 0; i items.length; i) {Item curr items[i];if (n curr.weight) {n - curr.weight;sum curr.value;}}return sum;} }得到的结果最大价值结果是1001630 实际上选择钻石和红宝石 得到的价值结果 1002400 贪心算法局限性