contest.cpp

/*
100633. Find Closest Person
User Accepted:89
User Tried:99
Total Accepted:90
Total Submissions:101
Difficulty:Easy
You are given three integers x, y, and z, representing the positions of three people on a number line:

x is the position of Person 1.
y is the position of Person 2.
z is the position of Person 3, who does not move.
Both Person 1 and Person 2 move toward Person 3 at the same speed.

Determine which person reaches Person 3 first:

Return 1 if Person 1 arrives first.
Return 2 if Person 2 arrives first.
Return 0 if both arrive at the same time.
Return the result accordingly.

 

Example 1:

Input: x = 2, y = 7, z = 4

Output: 1

Explanation:

Person 1 is at position 2 and can reach Person 3 (at position 4) in 2 steps.
Person 2 is at position 7 and can reach Person 3 in 3 steps.
Since Person 1 reaches Person 3 first, the output is 1.

Example 2:

Input: x = 2, y = 5, z = 6

Output: 2

Explanation:

Person 1 is at position 2 and can reach Person 3 (at position 6) in 4 steps.
Person 2 is at position 5 and can reach Person 3 in 1 step.
Since Person 2 reaches Person 3 first, the output is 2.

Example 3:

Input: x = 1, y = 5, z = 3

Output: 0

Explanation:

Person 1 is at position 1 and can reach Person 3 (at position 3) in 2 steps.
Person 2 is at position 5 and can reach Person 3 in 2 steps.
Since both Person 1 and Person 2 reach Person 3 at the same time, the output is 0.

 

Constraints:

1 <= x, y, z <= 100
*/

class Solution {
public:
    int findClosest(int x, int y, int z) {
        int dist1 = abs(x - z); // 第一个人到Person 3的距离
        int dist2 = abs(y - z); // 第二个人到Person 3的距离
        
        if (dist1 < dist2) return 1;
        if (dist2 < dist1) return 2;
        return 0;
    }
};