[Leetcode] 503. Next Greater Element II 解题报告

题目：

Given a circular array (the next element of the last element is the first element of the array), print the Next Greater Number for every element. The Next Greater Number of a number x is the first greater number to its traversing-order next in the array, which
means you could search circularly to find its next greater number. If it doesn't exist, output -1 for this number.

Example 1:

Input: [1,2,1]
Output: [2,-1,2]
Explanation: The first 1's next greater number is 2;
The number 2 can't find next greater number;
The second 1's next greater number needs to search circularly, which is also 2.

Note: The length of given array won't exceed 10000.
思路：

和[Leetcode] 496. Next Greater Element I比较类似，也会用到栈，只不过我们需要扫描两遍：第一遍可以找出所有greater element位于该数右侧的数；第二遍我们从左边再次进行扫描，如果发现当前数比栈顶元素大，那么就更新栈顶元素的greater element，并且出栈，直到栈顶元素不小于当前数。此时说明当前数一定是整个数组中最大的了，所以我们将当前数的greater
element置为-1即可。完成第二遍扫描之后，整个数组的greater element就全部找出来了。该算法的时间复杂度是O(n)，空间复杂度也是O(n)。

代码：

class Solution {
public:
vector<int> nextGreaterElements(vector<int>& nums) {
vector<int> ret(nums.size(), INT_MIN);
stack<pair<int, int>> st;
for (int i = 0; i < nums.size(); ++i) { // update the ones whose greater numbers are on their right
while (!st.empty() && st.top().first < nums[i]) {
pair<
4000
;int, int> p = st.top();
st.pop();
ret[p.second] = nums[i];
}
st.push(make_pair(nums[i], i));
}
for (int i = 0; i < nums.size(); ++i) { // update the ones whose greater numbers are on their left
while (!st.empty() && st.top().first < nums[i]) {
pair<int, int> p = st.top();
st.pop();
ret[p.second] = nums[i];
}
for (; ret[i] == INT_MIN; ++i) {
ret[i] = -1;
}
}
return ret;
}
};