739. Daily Temperatures

Given an array of integers temperatures represents the daily temperatures, return an array answer such that answer[i] is the number of days you have to wait after the ith day to get a warmer temperature. If there is no future day for which this is possible, keep answer[i] == 0 instead.

题目解析：

• 找右边第一个比它大的使用单调递减栈

代码：

class Solution {
    public int[] dailyTemperatures(int[] temperatures) {
        Deque<Integer> stack = new LinkedList<>();
        int[] result = new int[temperatures.length];
        for (int i = 0; i < temperatures.length; i++) {
            while(stack.size() > 0 && temperatures[stack.peek()] < temperatures[i]) {
                int idx = stack.pop();
                result[idx] = i - idx;
            }
            stack.push(i);
        }
        return result;
    }
}

496. Next Greater Element I

The next greater element of some element x in an array is the first greater element that is to the right of x in the same array.

You are given two distinct 0-indexed integer arrays nums1 and nums2, where nums1 is a subset of nums2.

For each 0 <= i < nums1.length, find the index j such that nums1[i] == nums2[j] and determine the next greater element of nums2[j] in nums2. If there is no next greater element, then the answer for this query is -1.

Return an array ans of length nums1.length such that ans[i] is the next greater element as described above.

题目解析：

• 先求出nums2中每个元素的下一个更大值

代码：

class Solution {
    public int[] nextGreaterElement(int[] nums1, int[] nums2) {
        Map<Integer, Integer> map = new HashMap<>();
        Deque<Integer> stack = new LinkedList<>();
        for (int i = 0; i < nums2.length; i++) {
            while(!stack.isEmpty() && nums2[i] > stack.peekLast()) {
                map.put(stack.pollLast(), nums2[i]);
            }
            stack.addLast(nums2[i]);
        }
        while(!stack.isEmpty()) {
            map.put(stack.pollLast(), -1);
        }
        int[] result = new int[nums1.length];
        for (int i = 0; i < nums1.length; i++) {
            result[i] = map.get(nums1[i]);
        }
        return result;
    }
}