LWC 61：739. Daily Temperatures

用户1147447

发布于 2018-01-02 11:00:31

4300

发布于 2018-01-02 11:00:31

文章被收录于专栏：机器学习入门

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

本文参与腾讯云自媒体同步曝光计划，分享自作者个人站点/博客。

原始发表：2017-12-04 ，如有侵权请联系 cloudcommunity@tencent.com 删除

java