【重点】【单调栈】496.下一个更大元素I

题目

法1：单调栈

Python

参考：本题题解
视频讲解

class Solution:
    def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]:
        val_idx_dict = {x: i for i, x in enumerate(nums1)}
        res = [-1] * len(nums1)
        stack = []
        for x in reversed(nums2):
            while stack and stack[-1] <= x:
                stack.pop()
            if stack and x in val_idx_dict:
                res[val_idx_dict[x]] = stack[-1]
            stack.append(x)
        return res

Java

class Solution {
    public int[] nextGreaterElement(int[] nums1, int[] nums2) {
        Map<Integer, Integer> map = new HashMap<>();
        Stack<Integer> stack = new Stack<>();
        for (int i = nums2.length - 1; i >= 0; --i) {
            while (!stack.isEmpty() && nums2[i] >= stack.peek()) {
                stack.pop();
            }
            map.put(nums2[i], stack.isEmpty() ? -1 : stack.peek());
            stack.push(nums2[i]);
        }

        int[] res = new int[nums1.length];
        for (int i = 0; i < nums1.length; ++i) {
            res[i] = map.get(nums1[i]);
        }
        return res;
    }
}

法2：迭代

最简单的方法，必须掌握！！！O(MN)

class Solution {
    public int[] nextGreaterElement(int[] nums1, int[] nums2) {
        int[] res = new int[nums1.length];
        Arrays.fill(res, -1);
        for (int i = 0; i < nums1.length; ++i) {
            int val = nums1[i];
            int nextBig = -1;
            for (int k = nums2.length - 1; k >= 0; --k) {
                if (nums2[k] == val) {
                    break;
                }
                if (nums2[k] > val) {
                    res[i] = nums2[k];
                }
            }
        }

        return res;
    }
}