(Easy) Reach a Number - LeetCode

标签：amp   exp   oal   ++   tput   turn   ==   ati   tin   

Description:

You are standing at position 0 on an infinite number line. There is a goal at position target.

On each move, you can either go left or right. During the n-th move (starting from 1), you take n steps.

Return the minimum number of steps required to reach the destination.

Example 1:

Input: target = 3
Output: 2
Explanation:
On the first move we step from 0 to 1.
On the second step we step from 1 to 3.

Example 2:

Input: target = 2
Output: 3
Explanation:
On the first move we step from 0 to 1.
On the second move we step  from 1 to -1.
On the third move we step from -1 to 2.

Note:

• target will be a non-zero integer in the range [-10^9, 10^9].

Solution:

class Solution {
    public int reachNumber(int target) {
        if(target <0){
              target = -target;
        }
        
        if(target ==0){
            return 0 ;   
        }
        int closest_i =0;
        for(int i = 1; (1+i)*i/2<target; i++){
         
            closest_i = i;
        }
      
        int min = closest_i+1;
       
        while(fun(min)%2 != target%2)
            min = min+1;
     
        return min;  
    }
    
    public int fun(int i ){
        
        return (i+1)*i/2;
    }
}

(Easy) Reach a Number - LeetCode

标签：amp   exp   oal   ++   tput   turn   ==   ati   tin   

原文地址：https://www.cnblogs.com/codingyangmao/p/11441488.html