LWC 61：739. Daily Temperatures

LWC 61：739. Daily Temperatures

传送门：739. Daily Temperatures

Problem:

Given a list of daily temperatures, produce a list that, for each day in the input, tells you how many days you would have to wait until a warmer temperature. If there is no future day for which this is possible, put 0 instead.

For example, given the list temperatures = [73, 74, 75, 71, 69, 72, 76, 73], your output should be [1, 1, 4, 2, 1, 1, 0, 0].

Note:

• The length of temperatures will be in the range [1, 30000]. Each temperature will be an integer in the range [30, 100].

思路：
求出当前元素的第一个大于它的位置即可，所以该问题就是nextGreaterElement系列，具体参考博文【 算法细节系列（10）：503. Next Greater Element II】

Java版本：

    public int[] dailyTemperatures(int[] temperatures) {        int n = temperatures.length;        int[] ans = new int[n];        Stack<Integer> stack = new Stack<>();        Map<Integer, Integer> mem = new HashMap<>();        for (int i = 0; i < n; ++i) {            while (!stack.isEmpty() && temperatures[stack.peek()] < temperatures[i]) {                int nxt = stack.pop();                mem.put(nxt, i);            }            stack.push(i);        }        for (int i = 0; i < n; ++i) {            if (mem.containsKey(i)) {                ans[i] = mem.get(i) - i;            }            else ans[i] = 0;        }        return ans;    }

Python 版本：

    def dailyTemperatures(self, temperatures):        """        :type temperatures: List[int]        :rtype: List[int]        """        n = len(temperatures)        stack = []        map = {}        ans = [0] * n        for i in range(n):            while (len(stack) > 0 and temperatures[stack[-1]] < temperatures[i]):                nxt = stack.pop(-1)                ans[nxt] = i - nxt            stack.append(i)        return ans