-
Notifications
You must be signed in to change notification settings - Fork 0
Expand file tree
/
Copy path0946_valide_stack_sequences.java
More file actions
78 lines (63 loc) · 2.36 KB
/
0946_valide_stack_sequences.java
File metadata and controls
78 lines (63 loc) · 2.36 KB
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
/*
* https://leetcode-cn.com/problems/validate-stack-sequences/
给定 pushed 和 popped 两个序列，只有当它们可能是在最初空栈上进行的推入 push 和弹出 pop 操作序列的结果时，返回 true；否则，返回 false 。
示例 1：
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：
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. pushed 是 popped 的排列。
-------------------------------------------------------------------------------------------------------------------------------
Given two sequences pushed and popped with distinct values, return true if and only if this could have been the result of a sequence of push and pop operations on an initially empty stack.
Example 1:
Input: pushed = [1,2,3,4,5], popped = [4,5,3,2,1]
Output: true
Explanation: We might do the following sequence:
push(1), push(2), push(3), push(4), pop() -> 4,
push(5), pop() -> 5, pop() -> 3, pop() -> 2, pop() -> 1
Example 2:
Input: pushed = [1,2,3,4,5], popped = [4,3,5,1,2]
Output: false
Explanation: 1 cannot be popped before 2.
Note:
1. 0 <= pushed.length == popped.length <= 1000
2. 0 <= pushed[i], popped[i] < 1000
3. pushed is a permutation of popped.
4. pushed and popped have distinct values.
*/
class Solution1 {
public boolean validateStackSequences(int[] pushed, int[] popped) {
int size = 0;
for (int i = 0, j = 0; i < pushed.length; i++) {
pushed[size++] = pushed[i];
while (size > 0 && pushed[size - 1] == popped[j]) {
size--;
j++;
}
}
return size == 0;
}
}
class Solution2 {
public boolean validateStackSequences(int[] pushed, int[] popped) {
int N = pushed.length;
Stack<Integer> stack = new Stack();
int j = 0;
for (int x: pushed) {
stack.push(x);
while (!stack.isEmpty() && j < N && stack.peek() == popped[j]) {
stack.pop();
j++;
}
}
return j == N;
}
}