Question

我有这个函数，它根据阵列分区的位置P返回数组两部分之和的最小差值。测试编程以在O（N * N）时间复杂度和0％性能下运行，尽管预期O（N）。

问题：我有什么方面可以改变以提高性能吗？有没有更好的方法在不使用子循环的情况下在循环内求和数组？感谢

任何整数P，使得0 <0。 P＆lt; N，将此磁带分成两个非空部分： A [0]，A [1]，...，A [P-1]和A [P]，A [P + 1]，...，A [N-1]。

两部分之间的差异是： |（A [0] + A [1] + ... + A [P-1]） - （A [P] + A [P + 1] + ... + A [N - 1]）|

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)) ;

Answer 1

是的，用线性时间（甚至是恒定的空间）和运行总和来做这件事是非常微不足道的。

function solution(arr) {
    var leftSum = 0; // sum from 0 to P
    var rightSum = arr.reduce((a, b) => a + b, 0); // sum from P to N
    var min = Math.abs(leftSum - rightSum); // initial value for p=0
    for (var p = 0; p < arr.length; p++)
        // move the element from the right to the left side
        leftSum += arr[p];
        rightSum -= arr[p];
        // then update minimum difference
        min = Math.min(min, Math.abs(leftSum - rightSum));
    }
    return min;
}

如何在时间复杂度为大O（N）的循环内对数组部分求和

1 个答案: