Daily Temperatures

问题描述（难度中等）

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].

方法一：Force遍历

两个循环，时间复杂度O(N^2)，空间复杂度O(1)。

package P739;

/**
 * 时间复杂度O(N^2)
 * 空间复杂度O(1)
 * 两边循环
 */
class Solution {
    public int[] dailyTemperatures(int[] T) {
        int[] result=new int[T.length];
        for (int x = 0; x < T.length; x++) {
            for (int y = x+1; y < T.length; y++) {
                if (T[x]<T[y]) {
                    result[x]=y-x;
                    break;
                }
            }
        }
        return result;
    }

    public static void main(String[] args) {
        int[] ints={73, 74, 75, 71, 69, 72, 76, 73};
        new Solution().dailyTemperatures(ints);
    }
}

方法二：Using Stack

暴力遍历的过程中我们发现有重复没有意义的比较，造成了N^2的复杂度。用Stack可以减少时间复杂度到O(N)，空间复杂度O(N)。

package P739;

import java.util.Stack;

/**
 * 时间复杂度O(N)
 * 空间复杂度O(N)
 * using stack
 * @autor yeqiaozhu.
 * @date 2019-10-25
 */
public class UsingStack {
    public int[] dailyTemperatures(int[] T) {
        int[] result=new int[T.length];
        Stack<Integer> stack=new Stack<>();
        for (int i = 0; i < T.length; i++) {
            while (!stack.isEmpty() && T[i]>T[stack.peek()]){
                int frontIndex=stack.pop();
                result[frontIndex]=i-frontIndex;
            }
            stack.push(i);
        }
        return result;
    }

    public static void main(String[] args) {
        int[] ints={73, 74, 75, 71, 69, 72, 76, 73};
        new UsingStack().dailyTemperatures(ints);
    }
}

总结

可以用栈减少时间复杂度，数组类的题目碰上重复的计算比较，即可考虑优化。