[LeetCode] 739. Daily Temperatures 每日温度

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].

给一个每天温度的数组，对于每一天，找出下一个比当前温度大的元素，记录它们之间的天数。

解法：栈，递减栈Descending Stack，新建一个长度和T相等的数组res，来记录天数。遍历数组，如果栈为空，直接如栈。如果栈不为空，且当前数字大于栈顶元素，pop出栈顶元素，求出下标差，也就是升温的天数，把这个差值记录给栈顶元素在res中的位置。然后继续看新的栈顶元素，直到当前数字小于等于栈顶元素停止。然后将当前数字入栈。最后返回res。

Java: Stack

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public int[] dailyTemperatures(int[] temperatures) {     Stack<Integer> stack = new Stack<>();     int[] ret = new int[temperatures.length];     for(int i = 0; i < temperatures.length; i++) {         while(!stack.isEmpty() && temperatures[i] > temperatures[stack.peek()]) {             int idx = stack.pop();             ret[idx] = i - idx;         }         stack.push(i);     }     return ret; }

Java: Array

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public int[] dailyTemperatures(int[] temperatures) {     int[] stack = new int[temperatures.length];     int top = -1;     int[] ret = new int[temperatures.length];     for(int i = 0; i < temperatures.length; i++) {         while(top > -1 && temperatures[i] > temperatures[stack[top]]) {             int idx = stack[top--];             ret[idx] = i - idx;         }         stack[++top] = i;     }     return ret; }

Python:

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# Time:  O(n) # Space: O(n)   class Solution(object):     def dailyTemperatures(self, temperatures):         """         :type temperatures: List[int]         :rtype: List[int]         """         result = [0] * len(temperatures)         stk = []         for i in xrange(len(temperatures)):             while stk and \                   temperatures[stk[-1]] < temperatures[i]:                 idx = stk.pop()                 result[idx] = i-idx             stk.append(i)         return result

Python: wo

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class Solution(object):     def dailyTemperatures(self, T):         """         :type T: List[int]         :rtype: List[int]         """         st = []         res = [0] * len(T)         for i in xrange(len(T)):             if not st or T[i] <= T[st[-1]]:                 st.append(i)             else:                 while st and T[i] > T[st[-1]]:                     j = st.pop()                     res[j] = i - j                 st.append(i)                                                     return res

C++:

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// Time:  O(n) // Space: O(n)   class Solution { public:     vector<int> dailyTemperatures(vector<int>& temperatures) {         vector<int> result(temperatures.size());         stack<int> stk;         for (int i = 0; i < temperatures.size(); ++i) {             while (!stk.empty() &&                    temperatures[stk.top()] < temperatures[i]) {                 const auto idx = stk.top(); stk.pop();                 result[idx] = i - idx;             }             stk.emplace(i);         }         return result;     } };

　　　　

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