LintCode 1060: Daily Temperatures (单调栈经典题)

原创已于 2023-09-16 12:13:47 修改 · 214 阅读

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#algorithm #算法

于 2020-08-24 15:34:08 首次发布

本文介绍了一种使用单调栈数据结构解决气温预测问题的方法。通过遍历每日温度列表，算法能高效地找出每一天之后首次出现的更暖温度，并返回等待天数列表。示例包括具体输入输出及代码实现。

1060. Daily Temperatures

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].

Example

Example 1:

Input: temperatures = [73, 74, 75, 71, 69, 72, 76, 73]

Output: [1, 1, 4, 2, 1, 1, 0, 0]

Explanation:

Just find the first day after it which has higher temperatures than it.

Example 2:

Input: temperatures = [50, 40, 39, 30]

Output: [0,0,0,0]

Notice

1.The length of temperatures will be in the range [1, 30000]. Each temperature will be an integer in the range [30, 100]

解法1：单调栈。

因为是对每个节点求右边第一个比其大的元素，所以用单调递减栈。

class Solution {
public:
    /**
     * @param temperatures: a list of daily temperatures
     * @return: a list of how many days you would have to wait until a warmer temperature
     */
    vector<int> dailyTemperatures(vector<int> &temperatures) {
        int n = temperatures.size();
        vector<int> result(n, 0);
        stack<int> s;
        for (int i = 0; i < n; ++i) {
            while(!s.empty() && temperatures[s.top()] < temperatures[i]) {
                result[s.top()] = i - s.top();
                s.pop();
            }
            s.push(i);
        }
        while(!s.empty()) {
            result[s.top()] = 0;
            s.pop();
        }
        
        return result;
    }
};

二刷：

class Solution {
public:
    vector<int> dailyTemperatures(vector<int>& temperatures) {
        int n = temperatures.size();
        stack<int> stk;
        vector<int> res(n, 0);
        for (int i = 0; i < n; i++) {
            while (!stk.empty() && temperatures[stk.top()] < temperatures[i]) {
                res[stk.top()] = i - stk.top();
                stk.pop();
            }
            stk.push(i);
        }
        return res;
    }
};

也可以反过来(n-1 .. 0)入栈。此时从栈底到栈顶仍然是递减，但如果按index顺序来则是递增。

vector<int> dailyTemperatures(vector<int>& temperatures) {
    int n = temperatures.size();
    vector<int> res(n);
    stack<int> stk; 
    for (int i = n - 1; i >= 0; i--) {
        while (!stk.empty() && temperatures[stk.top()] <= temperatures[i]) {
            stk.pop();
        }
        res[i] = stk.empty() ? 0 : (stk.top() - i); 
        stk.push(i); 
    }
    return res;
}