#### QUESTION:

There are n rings and each ring is either red, green, or blue. The rings are distributed across ten rods labeled from 0 to 9.

You are given a string rings of length 2n that describes the n rings that are placed onto the rods. Every two characters in rings forms a color-position pair that is used to describe each ring where:

The first character of the ith pair denotes the ith ring’s color (‘R’, ‘G’, ‘B’). The second character of the ith pair denotes the rod that the ith ring is placed on (‘0’ to ‘9’). For example, “R3G2B1” describes n == 3 rings: a red ring placed onto the rod labeled 3, a green ring placed onto the rod labeled 2, and a blue ring placed onto the rod labeled 1.

Return the number of rods that have all three colors of rings on them.

**Example 1:**

```
Input: rings = "B0B6G0R6R0R6G9"
Output: 1
Explanation:
- The rod labeled 0 holds 3 rings with all colors: red, green, and blue.
- The rod labeled 6 holds 3 rings, but it only has red and blue.
- The rod labeled 9 holds only a green ring.
Thus, the number of rods with all three colors is 1.
```

**Example 2:**

```
Input: rings = "B0R0G0R9R0B0G0"
Output: 1
Explanation:
- The rod labeled 0 holds 6 rings with all colors: red, green, and blue.
- The rod labeled 9 holds only a red ring.
Thus, the number of rods with all three colors is 1.
```

**Example 3:**

```
Input: rings = "G4"
Output: 0
Explanation:
Only one ring is given. Thus, no rods have all three colors.
```

**Constraints:**

```
rings.length == 2 * n
1 <= n <= 100
rings[i] where i is even is either 'R', 'G', or 'B' (0-indexed).
rings[i] where i is odd is a digit from '0' to '9' (0-indexed).
```

#### EXPLANATION:

当看到题目时, 第一个想法肯定是遍历循环, 然后计算出每个road上有几个rings. 这个思路是没错的. 然后再发现此时有可能会有同样颜色的ring出现在同一个road, 这里就需要做一个去重. 那么怎么样进行去重, 进行判断也行, 但是如果我们看过很多状态机的设定就会知道, 如果用位运算的话会简单很多. 比如第一位是代表红色. 第二位绿色. 那么直接或就可以拿到该road对应的值.

最后再判断当前road的3个bit位是否都为1, 也就是int值的7, 就可以判断出3个未知是否都有灯了.

#### SOLUTION:

```
class Solution {
func countPoints(_ rings: String) -> Int {
var ringsArray = Array(rings)
var roadsArray:[Int] = Array(repeating: 0, count: 10)
var r:Int = 0b1
var g:Int = 0b10
var b:Int = 0b100
for index in stride(from: 0, to: ringsArray.count-1, by: 2) {
var light = 0
switch ringsArray[index] {
case "R":
light = r
case "G":
light = g
case "B":
light = b
default:
light = 0
}
var road = Int(String(ringsArray[index+1]))
roadsArray[road!] = roadsArray[road!]|light
}
var result:Int = 0
for road in roadsArray {
if road == 7 {
result += 1
}
}
return result
}
}
```