源码记录

(1)类似正则表达式匹配问题， 例如s1 = "xxxxabxxxxbxx", s2 = "ab*b??"

public class Regex {static boolean scan(String s1, String s2){for(int i=0; i<s1.length(); i++){int cur = i, j;for(j=0; j<s2.length()&&cur<s1.length(); j++){if(s1.charAt(cur)==s2.charAt(j)){cur++;}else if(s2.charAt(j)=='?'){cur++;}else if(s2.charAt(j)=='*'){String tmps1 = s1.substring(cur), tmps2 = s2.substring(++j);boolean b = scan(tmps1, tmps2);if(b) return b;else break;}else if(s1.charAt(cur)!=s2.charAt(j)){break;}}if(j==s2.length()) return true;}return false;}public static void main(String [] args){String s1 = "xxaxxbxabcb";String s2 = "ab*b??";System.out.println(scan(s1, s2));}}

(2)求数组集合子集递归和非递归求法。

public class FindAllSubSet {public static void findAllSubSet(int [] a){int n = a.length;for(int i=0; i<Math.pow(2, n); i++){int cur = i;for(int j=0; j<n; j++){if((cur&1)==1){System.out.print(a[j]+" ");}cur >>= 1;}System.out.println();}}static void findSubSet(int [] a, boolean [] aux, int cur){//System.out.println(cur);if(cur>=a.length){for(int i=0; i<aux.length; i++){if(aux[i])System.out.print(a[i]+" ");}System.out.println();return;}int tmp = cur;findSubSet(a, aux, ++tmp);aux[cur] = true;findSubSet(a, aux, tmp);aux[cur] = false;}public static void main(String [] args){int [] a = {1, 2, 3};//findAllSubSet(a);boolean [] aux = new boolean[a.length];findSubSet(a, aux, 0);}}

（3）按照老鼠体重升序排序， 然后按速度找最大降序子序列

import java.util.Arrays;import java.util.Comparator;import java.util.LinkedList;import java.util.List;import java.util.Stack;public class SortMices {static class Mice{int weight, speed;public Mice(int w, int s){this.weight = w;this.speed = s;}public String toString(){return "weight = "+this.weight+"  speed="+this.speed;}}static void findMiceSet(Mice[] ms){Arrays.sort(ms, new Comparator<Mice>(){public int compare(Mice m1, Mice m2){if(m1.weight>m2.weight) return 1;else if(m1.weight<m2.weight) return -1;else return 0;}});show(ms);List<Mice> list = new LinkedList<Mice>();for(int i=0; i<ms.length; i++){int j;for(j=0; j<list.size(); j++){if(ms[i].speed>list.get(j).speed) break;}if(j==list.size()) list.add(ms[i]);else list.set(j, ms[i]);}System.out.println(list.size());int pre = Integer.MIN_VALUE;int j = ms.length-1;Stack<Mice> results = new Stack<Mice>();for(int i=list.size()-1; i>=0; i--){while(j>=0){if(ms[j].speed>pre&&ms[j].speed<=list.get(i).speed){results.push(ms[j]);pre = ms[j].speed;j--;break;}j--;}}while(!results.empty()){Mice m = results.pop();System.out.println(m.weight+" , "+m.speed);}}static void show(Mice [] ms){for(int i=0; i<ms.length; i++)System.out.println(ms[i]);}public static void main(String [] args){Mice [] ms = {new Mice(2, 5), new Mice(9, 1), new Mice(8, 2), new Mice(5, 4), new Mice(3, 3)};show(ms);System.out.println("+++++++++++++++++++++++++++++");findMiceSet(ms);}}

（4）合法的鸡蛋方法

/** * N个鸡蛋放入M个篮子中， 篮子不能为空， 要满足， 对任意不大于N的数量， 能用若干个篮子中的鸡蛋的和表示 * 方法是用枚举所有可能的排列方法。 * @author Administrator * */public class PutEggs {/** * 确认篮子可以组合出所有不大于N的数量 * @param n * @param bs * @return */static boolean confirm(int n, int[] bs){boolean [] ns = new boolean[n+1];int ss = bs.length;for(int i=0; i<Math.pow(2, ss); i++){int cur = i;int sum = 0;for(int j=0; j<ss;j++){if((cur&1)==1){sum += bs[j];}cur >>=1;}ns[sum] = true;}for(int i=1; i<ns.length; i++){if(!ns[i]) return false;}return true;}static void show(boolean [] bs){for(int i=0; i<bs.length; i++){System.out.print(bs[i]+" ");}System.out.println();}static void show(int [] bs){for(int i=0; i<bs.length; i++){System.out.print(bs[i]+" ");}System.out.println();}/** * 回溯遍历 * @param N * @param eggs * @param buckets * @param cur */static void putEggs(int N, int eggs, int [] buckets, int cur){if(eggs==0){if(cur==buckets.length&&confirm(N, buckets)){show(buckets);}return;}else if(cur==buckets.length){return;}int tmp = cur + 1;for(int i=eggs; i>0; i--){buckets[cur] = i;putEggs(N, eggs-i, buckets, tmp);}buckets[cur] = 0;}public static void main(String [] args){int [] bs = new int[5];putEggs(9, 9, bs, 0);}}

（5）统计字符数

import java.util.HashMap;/** * 时间复杂度O(N)时间内， 统计String中字符出现的次数 * @author Administrator * */public class CountChars {static HashMap<Character, Integer> count(String s ){int [] cc = new int [256];for(char a : s.toCharArray()){cc[a] ++;}HashMap<Character, Integer> counts = new HashMap<Character, Integer>();for(int i=0; i<cc.length; i++){if(cc[i]>0){counts.put((char)i, cc[i]);}}return counts;}public static void main(String [] args){String s = "AAbbaacccces";HashMap<Character, Integer> c = count(s);for(char a: c.keySet()){System.out.println(a+"  : "+c.get(a));}}}

（6）输入：一个长度为n的整数数组input 
输出：一个长度为n的整数数组result，满足result[i] = input数组中除了input[i]之外所有数的乘积（假设不会溢出）。比如输入：input = {2,3,4,5}，输出result = {60,40,30,24} 

public class CalculateProduct {public static int [] cal(int [] a){int [] re = new int[a.length];int pr = 1;for(int i=0; i<a.length; i++){re[i] = pr;pr *= a[i];}pr = a[a.length-1];for(int i = a.length-2; i>=0; i--){re[i] *= pr;pr *= a[i];}return re;}public static void show(int [] a){for(int i=0; i<a.length; i++){System.out.println(a[i]);}}public static void main(String [] args){int [] a = {1, 2, 3, 4, 5};a = cal(a);show(a);}}

（7）有n个人排队， 每个人都有要求Request， 具体每个要求是希望排在某个人之前或者之后， 用类RequestItem表示。

例如有 1， 2， 3三个人， 1希望排在2之后3之前， 2希望排在1之前， 3希望排在1， 2之后。输出一个合理的排列

一个拓扑排序的应用

import java.util.HashMap;import java.util.LinkedList;import java.util.List;class RequestItem{String name;boolean front;}class Request{String name;List<RequestItem> items;}public class FindValidOrder {static List<String> validOrder(List<String> names, List<Request> requests){HashMap<String, List<String>> maps = new HashMap<String, List<String>>();for(String name: names){maps.put(name, new LinkedList<String>());}for(Request re: requests){for(RequestItem item: re.items){if(item.front){List<String> tmp = maps.get(item.name);if(!tmp.contains(re.name)){tmp.add(re.name);}}else{List<String> tmp = maps.get(re.name);if(!tmp.contains(item.name)){tmp.add(item.name);}}}}HashMap<String, Integer> counts = new HashMap<String, Integer>();for(String name: names){counts.put(name, 0);}for(String name: names){List<String> tmp = maps.get(name);for(String s: tmp){int tmpcount = counts.get(s)+1;counts.put(s, tmpcount);}}List<String> queue = new LinkedList<String>();while(!counts.isEmpty()){boolean found = false;String next = null;for(String name: counts.keySet()){if(counts.get(name)==0){next = name;found = true;}}if(!found) break;counts.remove(next);for(String s: maps.get(next)){int tmpcount = counts.get(s)-1;counts.put(s, tmpcount);}queue.add(next);}if(queue.size() == names.size()) return queue;return null;}public static void main(String [] args){List<String> names = new LinkedList<String>();names.add("1");names.add("2");names.add("3");List<Request> requests = new LinkedList<Request>();Request r = new Request();r.name = "1";r.items = new LinkedList<RequestItem>();RequestItem item = new RequestItem();item.name = "2";item.front = true;r.items.add(item);RequestItem item2 = new RequestItem();item2.name = "3";item2.front = false;r.items.add(item2);requests.add(r);Request r2 = new Request();r2.name = "2";r2.items = new LinkedList<RequestItem>();RequestItem item3 = new RequestItem();item3.name = "1";item3.front = false;r2.items.add(item3);requests.add(r);Request r3 = new Request();r3.name = "3";r3.items = new LinkedList<RequestItem>();RequestItem item4 = new RequestItem();item4.name = "2";item4.front = true;r3.items.add(item4);RequestItem item5 = new RequestItem();item5.name = "1";item5.front = true;r3.items.add(item5);requests.add(r3);List<String> order = validOrder(names, requests);System.out.println(order);}}