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Merge pull request #31 from xukaide77/master
jvm第四次作业
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173 changes: 56 additions & 117 deletions
173
...220084/xukai_coding/coding-common/src/main/java/org/xukai/coderising/stack/InfixExpr.java
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104 changes: 104 additions & 0 deletions
104
...20084/xukai_coding/coding-common/src/main/java/org/xukai/coderising/stack/InfixExpr2.java
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| Original file line number | Diff line number | Diff line change |
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| @@ -0,0 +1,104 @@ | ||
| package org.xukai.coderising.stack; | ||
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| import java.util.HashMap; | ||
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| public class InfixExpr2 { | ||
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| public static enum TypeToken { | ||
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| add('+',0),substruct('-',1),devide('/',3),plus('*',2); | ||
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| TypeToken(char token, Integer priority) { | ||
| this.token = token; | ||
| this.priority = priority; | ||
| } | ||
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| private char token; | ||
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| private Integer priority; | ||
| } | ||
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| String expr = null; | ||
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| public InfixExpr2(String expr) { | ||
| this.expr = expr; | ||
| } | ||
| //3*20+12*5-40/2 | ||
| public Double evaluate() { | ||
| HashMap<String, Integer> map = new HashMap<>(); | ||
| map.put("+",0); | ||
| map.put("-",1); | ||
| map.put("/",3); | ||
| map.put("*",2); | ||
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| Stack tokenStack = new Stack(); | ||
| Stack numStack = new Stack(); | ||
| char[] chars = expr.toCharArray(); | ||
| boolean isNem = true; | ||
| for (int i = 0; i < expr.length(); i++) { | ||
| String token = (expr.charAt(i)) + ""; | ||
| Integer priprity = map.get(token); | ||
| if (priprity != null) { | ||
| //表示是运算符 | ||
| if (!tokenStack.isEmpty() && priprity < map.get(tokenStack.peek())) { | ||
| Float num2 = Float.valueOf(numStack.pop().toString()); | ||
| Float num1 = Float.valueOf(numStack.pop().toString()); | ||
| String pop = tokenStack.pop()+""; | ||
| if (pop.equals("-")) { | ||
| numStack.push(num1 - num2); | ||
| } else if (pop.equals("*")){ | ||
| numStack.push(num1 * num2); | ||
| } else if (pop.equals("/")){ | ||
| numStack.push(num1 / num2); | ||
| } else { | ||
| throw new RuntimeException(); | ||
| } | ||
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| } | ||
| tokenStack.push(token); | ||
| isNem = true; | ||
| } else if(token.matches("\\d{1}")) { | ||
| //表示是数字 | ||
| if (isNem) { | ||
| numStack.push(token); | ||
| } else { | ||
| numStack.push(numStack.pop().toString() + token); | ||
| } | ||
| isNem = false; | ||
| } else { | ||
| throw new RuntimeException(); | ||
| } | ||
| } | ||
| while (!tokenStack.isEmpty()) { | ||
| System.out.println(tokenStack.size()); | ||
| if (tokenStack.peek().equals("+")) { | ||
| Float num2 = Float.valueOf(numStack.pop().toString()); | ||
| Float num1 = Float.valueOf(numStack.pop().toString()); | ||
| numStack.push(num1+num2+""); | ||
| } else if (tokenStack.peek().equals("-")) { | ||
| Float num2 = Float.valueOf(numStack.pop().toString()); | ||
| Float num1 = Float.valueOf(numStack.pop().toString()); | ||
| numStack.push(num1-num2+""); | ||
| } else if (tokenStack.peek().equals("/")) { | ||
| Float num2 = Float.valueOf(numStack.pop().toString()); | ||
| Float num1 = Float.valueOf(numStack.pop().toString()); | ||
| numStack.push(num1/num2+""); | ||
| } else if (tokenStack.peek().equals("*")) { | ||
| Float num2 = Float.valueOf(numStack.pop().toString()); | ||
| Float num1 = Float.valueOf(numStack.pop().toString()); | ||
| numStack.push(num1*num2+""); | ||
| } else { | ||
| throw new RuntimeException(); | ||
| } | ||
| tokenStack.pop(); | ||
| } | ||
| // System.out.println(Double.valueOf(numStack.pop().toString())); | ||
| return Double.valueOf(numStack.pop().toString()); | ||
| } | ||
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| } |
87 changes: 87 additions & 0 deletions
87
.../xukai_coding/coding-common/src/main/java/org/xukai/coderising/stack/InfixToPostfix2.java
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,87 @@ | ||
| package org.xukai.coderising.stack; | ||
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| import java.util.List; | ||
| import java.util.Stack; | ||
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| /** | ||
| * 将中缀表达式转换为后缀表达式： | ||
| 与转换为前缀表达式相似，遵循以下步骤： | ||
| (1) 初始化两个栈：运算符栈S1和储存中间结果的栈S2； | ||
| (2) 从左至右扫描中缀表达式； | ||
| (3) 遇到操作数时，将其压入S2； | ||
| (4) 遇到运算符时，比较其与S1栈顶运算符的优先级： | ||
| (4-1) 如果S1为空，或栈顶运算符为左括号“(”，则直接将此运算符入栈； | ||
| (4-2) 否则，若优先级比栈顶运算符的高，也将运算符压入S1（注意转换为前缀表达式时是优先级较高或相同，而这里则不包括相同的情况）； | ||
| (4-3) 否则，将S1栈顶的运算符弹出并压入到S2中，再次转到(4-1)与S1中新的栈顶运算符相比较； | ||
| (5) 遇到括号时： | ||
| (5-1) 如果是左括号“(”，则直接压入S1； | ||
| (5-2) 如果是右括号“)”，则依次弹出S1栈顶的运算符，并压入S2，直到遇到左括号为止，此时将这一对括号丢弃； | ||
| (6) 重复步骤(2)至(5)，直到表达式的最右边； | ||
| (7) 将S1中剩余的运算符依次弹出并压入S2； | ||
| (8) 依次弹出S2中的元素并输出，结果的逆序即为中缀表达式对应的后缀表达式（转换为前缀表达式时不用逆序）。 | ||
| 例如，将中缀表达式“1+((2+3)×4)-5”转换为后缀表达式的过程如下： | ||
| 扫描到的元素 S2(栈底->栈顶) S1 (栈底->栈顶) 说明 | ||
| 1 1 空 数字，直接入栈 | ||
| + 1 + S1为空，运算符直接入栈 | ||
| ( 1 + ( 左括号，直接入栈 | ||
| ( 1 + ( ( 同上 | ||
| 2 1 2 + ( ( 数字 | ||
| + 1 2 + ( ( + S1栈顶为左括号，运算符直接入栈 | ||
| 3 1 2 3 + ( ( + 数字 | ||
| ) 1 2 3 + + ( 右括号，弹出运算符直至遇到左括号 | ||
| × 1 2 3 + + ( × S1栈顶为左括号，运算符直接入栈 | ||
| 4 1 2 3 + 4 + ( × 数字 | ||
| ) 1 2 3 + 4 × + 右括号，弹出运算符直至遇到左括号 | ||
| - 1 2 3 + 4 × + - -与+优先级相同，因此弹出+，再压入- | ||
| 5 1 2 3 + 4 × + 5 - 数字 | ||
| 到达最右端 1 2 3 + 4 × + 5 - 空 S1中剩余的运算符 | ||
| 因此结果为“1 2 3 + 4 × + 5 -”（注意需要逆序输出）。 | ||
| * @author xukai | ||
| * @desc | ||
| * @date 2017-04-22-14:10 | ||
| */ | ||
| public class InfixToPostfix2 { | ||
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| private static TokenParser tokenParser = new TokenParser(); | ||
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| public static String toPostFixExpr(String expr){ | ||
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| List<Token> tokens = tokenParser.parse(expr); | ||
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| Stack<Token> s1 = new Stack<Token>(); | ||
| Stack<Token> s2 = new Stack<Token>(); | ||
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| for (Token token : tokens) { | ||
| if (token.isNumber()) { | ||
| s2.push(token); | ||
| } else { | ||
| while (token.isOperator() && !s1.isEmpty()) { | ||
| if (!token.hasHigherPriority(s1.peek())) { | ||
| s2.push(s1.pop()); | ||
| continue; | ||
| } | ||
| break; | ||
| } | ||
| s1.push(token); | ||
| } | ||
| } | ||
| while (!s1.isEmpty()) { | ||
| s2.push(s1.pop()); | ||
| } | ||
| StringBuilder stringBuilder = new StringBuilder(); | ||
| while (!s2.isEmpty()) { | ||
| Token token = s2.pop(); | ||
| s1.push(token); | ||
| } | ||
| while (!s1.isEmpty()) { | ||
| Token token = s1.pop(); | ||
| stringBuilder.append(token.toString()).append(" "); | ||
| } | ||
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| return stringBuilder.substring(0,stringBuilder.length() - 1); | ||
| } | ||
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| } |
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