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  • LeetCode 739. Daily Temperatures

    原题链接在这里:https://leetcode.com/problems/daily-temperatures/

    题目:

    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].

    题解:

    Stack<Integer> stk里存index, 若当前temperature 更高, 就一直pop出来index, 计算天数差值保存.

    最后stk中剩余的index说明这些位置上没有更加暖和的天气, 存0.

    Time Complexity: O(n). n = T.length.

    Space: O(n).

    AC Java:

     1 class Solution {
     2     public int[] dailyTemperatures(int[] T) {
     3         if(T == null || T.length == 0){
     4             return T;
     5         }
     6         
     7         int len = T.length;
     8         int [] res = new int[len];
     9         Stack<Integer> stk = new Stack<Integer>();
    10         for(int i = len-1; i>=0; i--){
    11             while(!stk.isEmpty() && T[i]>=T[stk.peek()]){
    12                 stk.pop();
    13             }
    14             
    15             res[i] = stk.isEmpty() ? 0 : stk.peek()-i;
    16             
    17             stk.push(i);
    18         }
    19         
    20         return res;
    21     }
    22 }

    类似Next Greater Element I

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  • 原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/8192500.html
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