【算法】程序猿不写代码是不对的12

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package com.kingdz.algorithm.time201702;import java.util.ArrayList;import java.util.Arrays;import java.util.HashSet;import java.util.List;import java.util.Set;/** * 算数表达式分解<br/> * 程序的目的是将：(2+34)*5分解为List，其中分解后的结果为{(,2,+,34,),*,5}<br/> * 依然存在bug，就是分隔符自身含有包含的情况，比如分隔符集合中包含（和（（的情况。 *  * @author kingdz *  */public class Algo03 {/** * 分隔符集合 */public static final String[] separator = { "(", ")", "[", "]", "{", "}", "+", "-", "*", "/" };public static void main(String[] args) {String question = "(21+34)*5+12*34";List<String> result = dismantle(question);System.out.println(Arrays.toString(result.toArray()));}/** * 分解算法 *  * @param question * @return */private static List<String> dismantle(String question) {// 将字符串数组转换为set方便判断Set<String> separatorSet = new HashSet<String>();int maxLen = 0;for (String str : separator) {int len = str.length();if (len > maxLen) {maxLen = len;}separatorSet.add(str);}List<String> ret = new ArrayList<String>();// 确定开始和结尾来分解字符串int start = 0;int end = Math.min(start + maxLen, question.length());while (start < question.length() && start < end) {String sub = question.substring(start, end);// 是否包含分解字符串if (separatorSet.contains(sub)) {ret.add(sub);start = end;end = Math.min(start + maxLen, question.length());} else {if (end - start == 1) {if (ret.size() == 0) {// 如果集合为空则直接插入ret.add(sub);start = end;end = Math.min(start + maxLen, question.length());} else {// 如果不为空，则需要判断最后一位是否为分隔符String lastIn = ret.get(ret.size() - 1);if (separatorSet.contains(lastIn)) {ret.add(sub);} else {ret.set(ret.size() - 1, lastIn + sub);}start = end;end = Math.min(start + maxLen, question.length());}} else {// 将end逐渐较小，保证轮询所有长度的分割字符串数组end--;}}}return ret;}}
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