LeetCode/剑指 Offer 31. 栈的压入、弹出序列

剑指 Offer 31. 栈的压入、弹出序列

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

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。

提示:

  1. 0 <= pushed.length == popped.length <= 1000
  2. 0 <= pushed[i], popped[i] < 1000
  3. pushedpopped 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();
}
}

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();
}
}

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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