112th LeetCode Weekly Contest 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.
Note:
0 <= pushed.length == popped.length <= 1000
0 <= pushed[i], popped[i] < 1000
pushed
is a permutation ofpopped
.pushed
andpopped
have distinct values.
判断栈的合法性
class Solution { public: bool validateStackSequences(vector<int>& pushed, vector<int>& popped) { if(pushed.size()==0&&pushed.size()==0){ return true; } if(pushed.size()!=popped.size()){ return false; } stack<int>s; int index = 0,outdex = 0; while(index<pushed.size()&&outdex<=popped.size()){ if(pushed[index]==popped[outdex]){ index++; outdex++; }else if(!s.empty()&&popped[outdex]==s.top()){ while(!s.empty()&&popped[outdex]==s.top()){ s.pop(); outdex++; } }else{ s.push(pushed[index]); index++; } if (index >= popped.size())//如果入栈序列已经走完,出栈序列和栈顶元素一一比较 { while (!s.empty() && popped[outdex]==s.top()) { s.pop(); outdex++; } //如果和栈中比较完,两个序列都走完了,即表明顺序合法 if (index == pushed.size() && outdex == popped.size()) { return true; } else { return false; } } } } };
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