如何对时间复杂度为大O(N)的循环中的数组部分求和

我有一个函数,它根据数组被分割的位置P返回数组两部分之和之间的最小差。经测试,该程序的运行时间复杂度为O(N*N),性能为0%,但预期为O(N)。

问题:我是否可以在这方面进行任何更改以提高性能?有没有更好的方法可以不使用子循环而对循环中的数组求和?谢谢

任意整数P,0<<N、 将此磁带拆分为两个非空部分: A[0],A[1]。。。,A[P-1]和A[P],A[P+1]。。。,A[第1页]。

这两部分之间的差值为: |(A[0]+A[1]+…+A[P 1])(A[P]+A[P+1]+…+A[N 1])|

https://jsbin.com/mehisi/edit

function solution(A) {

  'use strict';

  if(arguments.length === 1 && typeof A === "object" && A.length > 1 ){
    try{
      const N = A.length;

      let diff;
      for( let P =1 ; P < N ; P++) {
        // For each value of P, calc the difference 
        //|(A[0] + A[1] + ... + A[P − 1]) − (A[P] + A[P + 1] + ... + A[N − 1])|

        // use slice to prevent modification of oraginal copy
        var A2 = A.slice(0) ; 
        //splice array into two A1 and A2
        let A1 = A2.splice(0, P);  // All Element from start up to P
        console.log("From Array " + A  + " Remove "+ A1 + " Remaining " + A2);
        // reduce((a, b) => a + b, 0); 
        let diffp = Math.abs((A1.reduce(function(a, b) { return a + b; }, 0)) - 
            (A2.reduce(function(a, b) { return a + b; }, 0))) ;

        if(diff > diffp || diff === undefined ){
          diff = diffp ;
        }
        console.log(P + "Difference ="+ diff + " Instead of " + diffp + " \r\n " );
      }

      // Return the Minimum value of P
      return diff  ;
    }
    catch(err){
     console.log("Error: " + err );
    return 0 ; // undefined ;
    }
  }else{
     console.log("Invalid parameter(s)");
    return 0 ; // undefined ;
  }

}

var A = [] ;
  A[0] = 5
  A[1] = 1
  A[2] = 2
  A[3] = 7
  A[4] = 4
console.log(solution(A)) ;