LeetCode679. 24 Game

You have 4 cards each containing a number from 1 to 9. You need to judge whether they could operated through *, /, +, -, (, ) to get the value of 24.

Example 1:

Input: [4, 1, 8, 7]
Output: True
Explanation: (8-4) * (7-1) = 24

Example 2:

Input: [1, 2, 1, 2]
Output: False

Note:

1. The division operator / represents real division, not integer division. For example, 4 / (1 - 2/3) = 12.
2. Every operation done is between two numbers. In particular, we cannot use - as a unary operator. For example, with [1, 1, 1, 1] as input, the expression -1 - 1 - 1 - 1 is not allowed.
3. You cannot concatenate numbers together. For example, if the input is [1, 2, 1, 2], we cannot write this as 12 + 12.

思路

有两种解法，其中一种不是很理解。先来说说另一种方法，Backtracking所有可能。对于给定的四个数，首先从中选两个数，然后对这两个数选择加减乘除这四种操作的一个，得出结果后从剩下的3个数中再选两个，执行一种操作。如此这样重复遍历所有可能即可。

class Solution {

    boolean res = false;
    final double eps = 0.001;

    public boolean judgePoint24(int[] nums) {
        List<Double> arr = new ArrayList<>();
        for(int n: nums) arr.add((double) n);
        helper(arr);
        return res;
    }

    private void helper(List<Double> arr){
        if(res) return;
        if(arr.size() == 1){
            if(Math.abs(arr.get(0) - 24.0) < eps)　// java中double类型判断是否相等的方法，不能直接用是否等于0判断
                res = true;
            return;
        }
        for (int i = 0; i < arr.size(); i++) {
            for (int j = 0; j < i; j++) {
                List<Double> next = new ArrayList<>();
                Double p1 = arr.get(i), p2 = arr.get(j);
                next.addAll(Arrays.asList(p1+p2, p1-p2, p2-p1, p1*p2));
                if(Math.abs(p2) > eps)  next.add(p1/p2);
                if(Math.abs(p1) > eps)  next.add(p2/p1);
                
                arr.remove(i);
                arr.remove(j);
                for (Double n: next){
                    arr.add(n);
                    helper(arr);
                    arr.remove(arr.size()-1);
                }
                arr.add(j, p2);
                arr.add(i, p1);
            }
        }
    }
}

 

注意上面List集合的两个方法：

void add(int index,
         E element)

Inserts the specified element at the specified position in this list (optional operation). Shifts the element currently at that position (if any) and any subsequent elements to the right (adds one to their indices).

---

E remove(int index)

Removes the element at the specified position in this list (optional operation). Shifts any subsequent elements to the left (subtracts one from their indices). Returns the element that was removed from the list.

Parameters:

index - the index of the element to be removed

Returns:

the element previously at the specified position

 

转载于:https://www.cnblogs.com/f91og/p/9503440.html