𝑰’𝒎 𝒉𝒉𝒈, 𝑰 𝒂𝒎 𝒂 𝒈𝒓𝒂𝒅𝒖𝒂𝒕𝒆 𝒔𝒕𝒖𝒅𝒆𝒏𝒕 𝒇𝒓𝒐𝒎 𝑵𝒂𝒏𝒋𝒊𝒏𝒈, 𝑪𝒉𝒊𝒏𝒂.
- 🏫 𝑺𝒉𝒄𝒐𝒐𝒍: 𝑯𝒐𝒉𝒂𝒊 𝑼𝒏𝒊𝒗𝒆𝒓𝒔𝒊𝒕𝒚
- 🌱 𝑳𝒆𝒂𝒓𝒏𝒊𝒏𝒈: 𝑰’𝒎 𝒄𝒖𝒓𝒓𝒆𝒏𝒕𝒍𝒚 𝒍𝒆𝒂𝒓𝒏𝒊𝒏𝒈 𝒅𝒆𝒔𝒊𝒈𝒏 𝒑𝒂𝒕𝒕𝒆𝒓𝒏, 𝑳𝒆𝒆𝒕𝒄𝒐𝒅𝒆, 𝒅𝒊𝒔𝒕𝒓𝒊𝒃𝒖𝒕𝒆𝒅 𝒔𝒚𝒔𝒕𝒆𝒎, 𝒎𝒊𝒅𝒅𝒍𝒆𝒘𝒂𝒓𝒆 𝒂𝒏𝒅 𝒔𝒐 𝒐𝒏.
- 💓 𝑯𝒐𝒘 𝒕𝒐 𝒓𝒆𝒂𝒄𝒉 𝒎𝒆：𝑽𝑿
- 📚 𝑴𝒚 𝒃𝒍𝒐𝒈: 𝒉𝒕𝒕𝒑𝒔://𝒉𝒉𝒈𝒚𝒚𝒅𝒔.𝒃𝒍𝒐𝒈.𝒄𝒔𝒅𝒏.𝒏𝒆𝒕/
- 💼 𝑷𝒓𝒐𝒇𝒆𝒔𝒔𝒊𝒐𝒏𝒂𝒍 𝒔𝒌𝒊𝒍𝒍𝒔：𝒎𝒚 𝒅𝒓𝒆𝒂𝒎
1-1: Description
给定一个整数数组 temperatures ，表示每天的温度，返回一个数组 answer ，其中 answer[i] 是指在第 i 天之后，才会有更高的温度。如果气温在这之后都不会升高，请在该位置用 0 来代替。
示例 1:
输入: temperatures = [73,74,75,71,69,72,76,73]
输出: [1,1,4,2,1,1,0,0]
示例 2:
输入: temperatures = [30,40,50,60]
输出: [1,1,1,0]
示例 3:
输入: temperatures = [30,60,90]
输出: [1,1,0]
提示：
1 <= temperatures.length <= 105
30 <= temperatures[i] <= 100
1-2: Idea - Monotone Stack
It isn’t the first time to encounter the monotone stack or queue. It is a pity that I have not to AC this question. So I record the problem resolving process in this article. A monotonically decreasing stack can be used to resolve these similar question.
The main idea to use a monotonically decreasing stack is that we should keep the order of the stack. When coming to a value, we should judge whether is bigger than the top value. If true, remove it and judge the next top value. The process of removing the top value is also used to find the first bigger value than which was removed. The next step is to calculate the space of the two values i.e. the two numbers being compared.
Another point to be taken into consideration is that instead of storing the value in the stack, we should store the index of each value. Why? Because we should record the space of the two values. If we record the indexes, we can calculate the space and get the value through the index value, right?
1-3: Demo
public class Solution {
public int[] dailyTemperatures(int[] temperatures) {
int n = temperatures.length;
int[] res = new int[n];
if (n <= 1) {
return res;
}
// record index
Deque<Integer> stack = new LinkedList<>();
for (int i = 0; i < n; i++) {
int temperature = temperatures[i];
while (!stack.isEmpty() && temperatures[stack.peek()] < temperature) {
int preIndex = stack.pop();
res[preIndex] = i - preIndex;
}
stack.push(i);
}
return res;
}
}
1-4: Conclusion
The monotone stack could be used to find the first bigger or smaller value than one in an array.