输入两个整数序列,第一个序列表示栈的压入顺序,请判断第二个序列是否为该栈的弹出顺序。假设压入栈的所有数字均不相等。例如,序列 {1,2,3,4,5} 是某栈的压栈序列,序列 {4,5,3,2,1} 是该压栈序列对应的一个弹出序列,但 {4,3,5,1,2} 就不可能是该压栈序列的弹出序列。
示例1:
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| 输入:pushed = [1,2,3,4,5], popped = [4,5,3,2,1] 输出:true 解释:我们可以按以下顺序执行: push(1), push(2), push(3), push(4), pop() -> 4, push(5), pop() -> 5, pop() -> 3, pop() -> 2, pop() -> 1
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示例2:
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| 输入:pushed = [1,2,3,4,5], popped = [4,3,5,1,2] 输出:false 解释:1 不能在 2 之前弹出。
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提示:
0 <= pushed.length == popped.length <= 1000
0 <= pushed[i], popped[i] < 1000
pushed 是 popped 的排列。
来源:力扣(LeetCode)
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题解:
本题只讲思路,至于是用数组、栈、还是链表实现由你。遍历pushed数组,如果当前元素和popped数组当前所指元素相等,则指向pushed数组指针回退一格,指向popped数组指针前进一格,如果不等则指向pushed数组指针继续向前遍历,这里我采用栈结构实现。
具体代码如下:
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| class Solution { public boolean validateStackSequences(int[] pushed, int[] popped) { Deque<Integer> stack = new ArrayDeque(); int popIndex = 0; for (int x : pushed) { stack.push(x); while (!stack.isEmpty() && stack.peek() == popped[popIndex]) { stack.pop(); popIndex++; } } return stack.isEmpty(); } }
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链表结构实现:
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| class Solution { public boolean validateStackSequences(int[] pushed, int[] popped) { LinkedList<Integer> stack = new LinkedList<>(); int popIndex = 0; for (int x : pushed) { stack.push(x); while (!stack.isEmpty() && stack.peek() == popped[popIndex]) { stack.pop(); popIndex++; } } return stack.isEmpty(); } }
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数组结构实现:
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| class Solution { public boolean validateStackSequences(int[] pushed, int[] popped) { int[] stack = new int[pushed.length]; int pushIndex = 0, popIndex = 0; for (int i = 0; i < pushed.length; i++) { stack[pushIndex++] = pushed[i]; while (pushIndex > 0 && popped[popIndex] == stack[pushIndex - 1]) { pushIndex--; popIndex++; } } return pushIndex == 0; } }
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