标签：cat rev point eal cts strong 遍历 als 0.00
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:
/ represents real division, not integer division. For example, 4 / (1 - 2/3) = 12.- as a unary operator. For example, with [1, 1, 1, 1] as input, the expression -1 - 1 - 1 - 1 is not allowed.[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)
E remove(int index)
index - the index of the element to be removed
标签：cat rev point eal cts strong 遍历 als 0.00
原文地址：https://www.cnblogs.com/f91og/p/9503440.html
