Class Stack<E>
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Method Summary
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?boolean
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empty
()
??????????Tests if this stack is empty.
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E
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peek
()
??????????Looks at the object at the top of this stack without removing it from the stack.
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E
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pop
()
??????????Removes the object at the top of this stack and returns that object as the value of this function.
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E
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push
(
E
?item)
??????????Pushes an item onto the top of this stack.
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?int
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search
(
Object?o)
??????????Returns the 1-based position where an object is on this stack.
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另外還繼承了java.util.Vector<E>中的基本方法，常見的用法如下所示：
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import java.util.*;
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public class kkk{
?public static void main(String args[]){
??int m = 1;
??Stack<Integer> a = new Stack<Integer>();
??
??a.push(m);?? //壓入1
??a.push(m+1); //壓入2
??a.push(m+2); //壓入3
??a.push(m+3); //壓入4
??a.push(m+4); //壓入5
??? System.out.println(a.push(m+5)); //壓入6,并返回當(dāng)前值
?
??? System.out.println(a.empty());?? //判斷棧是否為空,為空則TRUE
??? System.out.println(a.isEmpty()); //判斷棧是否為空,為空則TRUE
???
??System.out.println(a.size());???? //返回當(dāng)前棧長度
??System.out.println(a.toString()); //返回當(dāng)前棧的內(nèi)容
??
??System.out.println(a.lastElement());? //返回當(dāng)前棧中的最后一個(gè)元素6
??System.out.println(a.firstElement()); //返回當(dāng)前棧中的第一個(gè)元素1
??
??System.out.println(a.peek()); //返回當(dāng)前值6
??System.out.println(a.pop());? //返回當(dāng)前值并壓出6
??System.out.println(a.pop());? //返回當(dāng)前值并壓出5
??
??System.out.println(a.search(3));
??//搜索“3”在棧中出現(xiàn)的位置,頂端為1,其余依次累加,若有多個(gè)則返回第1個(gè)的位置
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??a.add(m+9); //壓入10，跟push一樣的效果
??a.clear(); //清空棧
??
??? System.out.println(a.empty());?? //判斷棧是否為空,為空則TRUE
??? System.out.println(a.isEmpty()); //判斷棧是否為空,為空則TRUE?
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?}
}
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-The End-