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;/** * 算數表達式分解<br/> * 程序的目的是將：(2+34)*5分解為List，其中分解后的結果為{(,2,+,34,),*,5}<br/> * 依然存在bug，就是分隔符自身含有包含的情況，比如分隔符集合中包含（和（（的情況。 *  * @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) {		// 將字符串數組轉換為set方便判斷		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>();		// 確定開始和結尾來分解字符串		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) {						// 如果集合為空則直接插入						ret.add(sub);						start = end;						end = Math.min(start + maxLen, question.length());					} else {						// 如果不為空，則需要判斷最后一位是否為分隔符						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 {					// 將end逐漸較小，保證輪詢所有長度的分割字符串數組					end--;				}			}		}		return ret;	}}