【leetcode】Reach a Number

题目:

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].

 

解题思路:

  看完题目后,我脑子里首先出现的是动态规划算法解决这一类问题。但是仔细想想,又觉得不太对,首先target的范围很大,没有这个大的数组可以保存中间结果。之后脑子里闪过了无数的方法,但都被一一否决了。万般无奈之下,想起了“找规律”的老办法。题目要求是从0开始,第n次操作可以到达target,那么可以先试试找出每次操作可以到达的的number,是否能够发现其中的规律。

      

function unique(a) {
    var res = [];

    for (var i = 0, len = a.length; i < len; i++) {
        var item = a[i];

        for (var j = 0, jLen = res.length; j < jLen; j++) {
            if (res[j] === item)
                break;
        }

        if (j === jLen)
            res.push(item);
    }

    return res;
}

var reachNumber = function(target) {
    var l = [0]
    for(var i = 1;i < 6;i++){ 
        var tl = []
        while(l.length > 0){
            var t = l.pop()
            tl.push(t-i)
            tl.push(t+i)
        }
        var tl = unique(tl).sort(function(a,b){
            return a-b})
        //console.log(i,':',tl[0],tl[1],tl[2],tl[3])
        console.log(i,':',tl)
        for(var j =0;j){
            l.push(tl[j])
        }
    }
};

reachNumber()

输出的结果如下:

1 ':' [ -1, 1 ]
2 ':' [ -3, -1, 1, 3 ]
3 ':' [ -6, -4, -2, 0, 2, 4, 6 ]
4 ':' [ -10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10 ]
5 ':' [ -15, -13, -11, -9, -7, -5, -3, -1, 1, 3, 5, 7, 9, 11, 13, 15 ]

为了控制输出的长度,这里设置了i<6的条件,其实适当把i放大,会发现更明显的规律。

这里就直接把规律列出来了:

a. 第n次操作能到达的最大范围是 -(1+2+...+n)和 (1+2+...+n);

b. 负数的number和正数的number是对称的,可以令target = abs(target);

c. n%4 == 0或者n%4 == 3的时候,只能移动到小于偶数的number;

d. n%4 == 1或者n%4 == 2的时候,只能移动到奇数的number;

所以,要找出最小的n,可以到达abs(target)分两种情况:

1. abs(target)是偶数,需要满足上面的a和c两个条件;

2.abs(target)是奇数,需要满足上面的a和d两个条件;

 

代码如下:

var reachNumber = function(target) {
    if (target < 0){
        target = -target
    }
    if(target == 1 || target == -1){
        return 1
    }
    var isOdd = target%2
    var count = 1;
    var t = 1;
    while(count++) {
        t += count
        if (t >= target && isOdd == 0 && (count % 4 == 3 || count % 4 == 0)) {
            return count;
        }
        else if (t >= target && isOdd != 0 && (count % 4 == 1 || count % 4 == 2)) {
            return count
        }

    }
};

console.log(reachNumber(1))

 

转载于:https://www.cnblogs.com/seyjs/p/8191963.html

你可能感兴趣的:(数据结构与算法)