按题意模拟即可：
一个指针cur依次指向target中的每个元素，另一个指针i依次指向1∼n的数字；
对i所指向的每个数字进行Push操作，然后判断当前数字与target[cur]是否相等；
Pop操作。过程中需注意cur的越界，当其越界则target构造完毕。
class Solution {
public List<String> buildArray(int[] target, int n) {
List<String> res = new ArrayList<>();
for (int i = 1, cur = 0; i <= n && cur < target.length; i++) {
res.add("Push");
if (target[cur] != i)
res.add("Pop");
else
cur++;
}
return res;
}
}
class Solution {
public:
vector<string> buildArray(vector<int>& target, int n) {
vector<string> res;
for (int i = 1, cur = 0; i <= n && cur < target.size(); i++) {
res.emplace_back("Push");
if (target[cur] != i)
res.emplace_back("Pop");
else
cur++;
}
return res;
}
};
impl Solution {
pub fn build_array(target: Vec<i32>, n: i32) -> Vec<String> {
let mut res = Vec::new();
let mut cur = 0;
for i in 1..(n + 1) {
if (cur < target.len()) {
res.push(String::from("Push"));
if (target[cur] != i) {
res.push(String::from("Pop"));
}
else {
cur += 1;
}
}
}
res
}
}
