StackOverflow2013

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15781419
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0
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2013-04-03T07:33:09.443
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2013-04-03T13:30:55.133
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2013-04-03T13:30:55.133
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text
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What you need to do if you really want as few parentheses as possible, is similar to what the algorithm you linked to says. However... <ul> <li>You should store an operator for each composite operand in the <code>Stack</code>. Namely, the last operator used in the operand. You could use a second <code>Stack</code> for this. If the operand is not composite, you could add <code>null</code> to the second <code>Stack</code>, since there is no operator.</li> <li>Don't encapsulate the resulted <code>String</code> with parentheses. That is done elsewhere in the algorithm (see below).</li> </ul> When you pop the top two values from each of the <code>Stack</code>s, you have 3 operators at hand: <ul> <li>The current operator</li> <li>The last used operator in the first operand (if the operator exists)</li> <li>The last used operator in the second operand (if the operator exists)</li> </ul> Depending on these three operators, you should encapsulate the first and/or second operand with parentheses, before combining them. You could use operator precedence to determine whether there should be parentheses. The order goes like this: <code>(none), {"*", "/"}, {"+", "-"}</code> <ul> <li>The first operand needs parentheses if and only if its operator has a lower precedence than the current operator.</li> <li>The second operand needs parentheses if its operator has a lower precedence than the current operator, or if they have equal precedence where the current operator is either <code>"/"</code> or <code>"-"</code>.</li> </ul> The rest should be done the way your algorithm described.
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1. POPostfix to Infix with minimum number of parentheses
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1. USLone nebula
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1. USLone nebula
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1. POPostfix to Infix with minimum number of parentheses
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1. COWhat about operator precedence?
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 USMichael Deardeuff
2. CO@Michael You are right. I edited it.
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