[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

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

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:

# 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

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++:

// 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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