946 Validate Stack Sequences

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.

Constraints:

• 0 <= pushed.length == popped.length <= 1000

• 0 <= pushed[i], popped[i] < 1000

• pushed is a permutation of popped.

• pushed and popped have distinct values.

这题之前想了半天，放弃了，看了答案，竟然是贪心。我们首先用个empty的栈，把push一个一个往里放，然后，如果我们看到现在栈顶等于我们pop的数的话，我们就贪心地把数字pop掉。因为如果不及时pop掉的话，再push以后就不能pop了。T:O(N), S:O(N)

public boolean validateStackSequences(int[] pushed, int[] popped) {
     if (pushed == null || popped == null || pushed.length != popped.length) {
         return false;
     }

    Stack<Integer> stack = new Stack<>();
    int j = 0;
    int popLen = popped.length;

    for (int pushVal : pushed) {
        stack.push(pushVal);
        while (!stack.isEmpty() && j < popLen && stack.peek() == popped[j]) {
            stack.pop();
            j++;
        }
    }

    return j == popLen;
}