Leetcode

# 946. Validate Stack Sequences

by hooni posted Apr 08, 2020
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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 Solution {
public boolean validateStackSequences(int[] pushed, int[] popped) {
int loop = pushed.length;
Stack<Integer> stack = new Stack();

int index = 0;
//for (int x: pushed) {
for(int i=0; i<loop; i++){
//stack.push(x);
stack.push(pushed[i]);
while (!stack.isEmpty() && index < loop && stack.peek() == popped[index]) {
stack.pop();
index++;
}
}

return index == loop;
}
}
```

```class Solution {
public boolean validateStackSequences(int[] pushed, int[] popped) {
int loop = pushed.length;
int[] stack = new int[loop];
int peek = -1;

int curr = 0;
for(int i=0; i<loop; i++){
stack[++peek] = pushed[i];

while(peek > -1 && curr < loop && stack[peek] == popped[curr]){
peek--;
curr++;
}
}

return curr == loop;
}
}
```

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