Description#
Given a circular integer array nums (i.e., the next element of nums[nums.length - 1] is nums[0]), return the next greater number for every element in nums.
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, return -1 for this number.
Example 1:
Input: nums = [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.
Example 2:
Input: nums = [1,2,3,4,3]
Output: [2,3,4,-1,4]
Constraints:
1 <= nums.length <= 104-109 <= nums[i] <= 109
Solutions#
Solution 1#
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| class Solution:
def nextGreaterElements(self, nums: List[int]) -> List[int]:
n = len(nums)
ans = [-1] * n
stk = []
for i in range(n << 1):
while stk and nums[stk[-1]] < nums[i % n]:
ans[stk.pop()] = nums[i % n]
stk.append(i % n)
return ans
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| class Solution {
public int[] nextGreaterElements(int[] nums) {
int n = nums.length;
int[] ans = new int[n];
Arrays.fill(ans, -1);
Deque<Integer> stk = new ArrayDeque<>();
for (int i = 0; i < (n << 1); ++i) {
while (!stk.isEmpty() && nums[stk.peek()] < nums[i % n]) {
ans[stk.pop()] = nums[i % n];
}
stk.push(i % n);
}
return ans;
}
}
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| class Solution {
public:
vector<int> nextGreaterElements(vector<int>& nums) {
int n = nums.size();
vector<int> ans(n, -1);
stack<int> stk;
for (int i = 0; i < (n << 1); ++i) {
while (!stk.empty() && nums[stk.top()] < nums[i % n]) {
ans[stk.top()] = nums[i % n];
stk.pop();
}
stk.push(i % n);
}
return ans;
}
};
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| func nextGreaterElements(nums []int) []int {
n := len(nums)
ans := make([]int, n)
for i := range ans {
ans[i] = -1
}
var stk []int
for i := 0; i < (n << 1); i++ {
for len(stk) > 0 && nums[stk[len(stk)-1]] < nums[i%n] {
ans[stk[len(stk)-1]] = nums[i%n]
stk = stk[:len(stk)-1]
}
stk = append(stk, i%n)
}
return ans
}
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| function nextGreaterElements(nums: number[]): number[] {
const stack: number[] = [],
len = nums.length;
const res: number[] = new Array(len).fill(-1);
for (let i = 0; i < 2 * len - 1; i++) {
const j = i % len;
while (stack.length !== 0 && nums[stack[stack.length - 1]] < nums[j]) {
res[stack[stack.length - 1]] = nums[j];
stack.pop();
}
stack.push(j);
}
return res;
}
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| /**
* @param {number[]} nums
* @return {number[]}
*/
var nextGreaterElements = function (nums) {
const n = nums.length;
let stk = [];
let ans = new Array(n).fill(-1);
for (let i = 0; i < n << 1; i++) {
const j = i % n;
while (stk.length && nums[stk[stk.length - 1]] < nums[j]) {
ans[stk.pop()] = nums[j];
}
stk.push(j);
}
return ans;
};
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Solution 2#
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| class Solution:
def nextGreaterElements(self, nums: List[int]) -> List[int]:
n = len(nums)
ans = [-1] * n
stk = []
for i in range(n * 2 - 1, -1, -1):
i %= n
while stk and stk[-1] <= nums[i]:
stk.pop()
if stk:
ans[i] = stk[-1]
stk.append(nums[i])
return ans
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| class Solution {
public int[] nextGreaterElements(int[] nums) {
int n = nums.length;
int[] ans = new int[n];
Arrays.fill(ans, -1);
Deque<Integer> stk = new ArrayDeque<>();
for (int i = n * 2 - 1; i >= 0; --i) {
int j = i % n;
while (!stk.isEmpty() && stk.peek() <= nums[j]) {
stk.pop();
}
if (!stk.isEmpty()) {
ans[j] = stk.peek();
}
stk.push(nums[j]);
}
return ans;
}
}
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| class Solution {
public:
vector<int> nextGreaterElements(vector<int>& nums) {
int n = nums.size();
vector<int> ans(n, -1);
stack<int> stk;
for (int i = n * 2 - 1; ~i; --i) {
int j = i % n;
while (!stk.empty() && stk.top() <= nums[j]) stk.pop();
if (!stk.empty()) ans[j] = stk.top();
stk.push(nums[j]);
}
return ans;
}
};
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| func nextGreaterElements(nums []int) []int {
n := len(nums)
ans := make([]int, n)
for i := range ans {
ans[i] = -1
}
var stk []int
for i := n*2 - 1; i >= 0; i-- {
j := i % n
for len(stk) > 0 && stk[len(stk)-1] <= nums[j] {
stk = stk[:len(stk)-1]
}
if len(stk) > 0 {
ans[j] = stk[len(stk)-1]
}
stk = append(stk, nums[j])
}
return ans
}
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| /**
* @param {number[]} nums
* @return {number[]}
*/
var nextGreaterElements = function (nums) {
const n = nums.length;
let stk = [];
let ans = new Array(n).fill(-1);
for (let i = n * 2 - 1; ~i; --i) {
const j = i % n;
while (stk.length && stk[stk.length - 1] <= nums[j]) {
stk.pop();
}
if (stk.length) {
ans[j] = stk[stk.length - 1];
}
stk.push(nums[j]);
}
return ans;
};
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