找出 int 数组的平衡点 & 二叉树 / 平衡二叉树 / 满二叉树 / 完全二叉树 / 二叉查找树

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mob604756ea4c07 2020-06-20 17:05:00

文章标签 平衡点二叉树算法 Algorithm js 文章分类 代码人生

找出 int 数组的平衡点树算法二叉树 / 平衡二叉树 / 满二叉树 / 完全二叉树 / 二叉查找树 balanced binary tree

找出 int 数组的平衡点

左右两边和相等, 若存在返回平衡点的值(可能由多个); 若不存在返回 -1;

``java

int [] arr = {2,3,4,2,4};


```js
const  arr = [2,3,4,2,4];

https://repl.it/@xgqfrms/find-number-array-balance-point

https://repl.it/@xgqfrms/find-Int-array-balance-point

// 找出 Int 数组平衡点

/**
 * 整形数组平衡点问题：平衡点指左边的整数和等于右边的整数和，
 * 求出平衡点位置，要求输入的数组可能是GB级
 *
 * 要求找出整型数组的一个平衡点（如果要找出所有平衡点的话，按此方法需要把每一个平衡点都存起来）
 */

const log = console.log;

// 'public' modifier cannot appear on a module or namespace element.ts
// public class IntArrayBalancePoint {

class IntArrayBalancePoint {
  constructor(args: String[]) {
    log(`args`, args)
    // const a: Object = [- 7 , 1, 5, 2, -5, 1];
    // const b: Object = [2, 3, 4, 2, 4];
    // const c: Object = [2, 3, 4, 3, 2];
    // const a: Number[] = [- 7 , 1, 5, 2, -5, 1];
    // const b: Number[] = [2, 3, 4, 2, 4];
    // const c: Number[] = [2, 3, 4, 3, 2];
    // interface Number
    // An object that represents a number of any kind.
    // All JavaScript numbers are 64-bit floating-point numbers.
    // const a: number[] = [- 7 , 1, 5, 2, -5, 1];
    // const b: number[] = [2, 3, 4, 2, 4];
    // const c: number[] = [2, 3, 4, 3, 2];
    // const t = new IntArrayBalancePoint([]);
    // log(t.findBalancePoint(a));
    // log(t.findBalancePoint(b));
    // log(t.findBalancePoint(c));
  }
  public findBalancePoint(a: number[]) {
  // findBalancePoint(a: number[]) {
    const len = a.length || 0;
    if (a === null) {
      return -1;
    }
    let sum = 0;
    let subSum = 0;
    for (let i = 0; i < len; i ++) {
        sum += a[i];
    }
    for (let i = 0; i < len; i++) {
        if (subSum === sum - subSum - a[i]) {
            // log(a[i]);
            return a[i];
        } else {
            subSum += a[i];
        }
    }
    return -1;
  }
}


const a: number[] = [- 7 , 1, 5, 2, -5, 1];
const b: number[] = [2, 3, 4, 2, 4];
const c: number[] = [2, 3, 4, 3, 2];

const t = new IntArrayBalancePoint([]);
log(t.findBalancePoint(a));
log(t.findBalancePoint(b));
log(t.findBalancePoint(c));

// args []
// -5
// -1
// 4

refs

java

package find_Int_array_balance_point;

// 找出 Int 数组平衡点

/**
* 整形数组平衡点问题：平衡点指左边的整数和等于右边的整数和，
* 求出平衡点位置，要求输入的数组可能是GB级
*
* 要求找出整型数组的一个平衡点（如果要找出所有平衡点的话，按此方法需要把每一个平衡点都存起来）
*/

public class IntArrayBalancePoint {
public static void main(String[] args) {
int[] a = { - 7 , 1, 5, 2, -5, 1} ;
int[] b = {2, 3, 4, 2, 4} ;
int[] c = {2, 3, 4, 3, 2} ;
IntArrayBalancePoint t = new IntArrayBalancePoint();
System.out.println(t.findBalancePoint(a));
System.out.println(t.findBalancePoint(b));
System.out.println(t.findBalancePoint(c));
// t.findBalancePoint(a);
// t.findBalancePoint(b);
// t.findBalancePoint(c);
  }
public int findBalancePoint(int[] a) {
if (a == null) {
return -1;
    }
long sum = 0l;
long subSum = 0l;
for ( int i = 0 ; i < a.length; i ++ ) {
sum += a[i];
    }
for (int i = 0; i < a.length; i ++ ) {
if (subSum == sum - subSum - a[i]) {
// System.out.println(a[i]);
return a[i];
        } else {
subSum += a[i];
        }
    }
return -1;
  }
}