LeetCode : 739. Daily Temperatures. 日常温度 素颜马尾好姑娘i 2021-06-24 16:11 225阅读 0赞 **试题** Given a list of daily temperatures T, return a list such 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 of temperatures T = \[73, 74, 75, 71, 69, 72, 76, 73\], 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 \[30, 100\]. **代码** 这利用了栈的一个非常好的特性,就是我们可以根据弹入的数据找到一个单调递增的数列。 当我们从后往前押入栈时,如果栈内数据小于当前数据,就可以直接弹出,这样我们就找到比当前数据大的第一个数据。 class Solution { public int[] dailyTemperatures(int[] T) { int[] res = new int[T.length]; Stack<Integer> sta = new Stack<Integer>(); for(int i=T.length-1; i>=0; i--){ while(!sta.isEmpty() && T[i]>=T[sta.peek()]) sta.pop(); res[i] = sta.isEmpty()?0:sta.peek()-i; sta.push(i); } return res; } }
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