JavaScript递归函数保留原数组获取原始索引的代码修改咨询

方案1：新增可选参数透传原始数组 #

给递归函数增加第四个可选参数存储原始数组，仅首次调用时初始化，递归时直接透传即可：

function subsetSum(numbers, target, partial, originalList) {
  var s, n, remaining;
  // 首次调用时如果没传originalList，默认用当前传入的完整数组作为原始数组
  originalList = originalList || numbers;
  partial = partial || [];

  // sum partial
  s = partial.reduce(function (a, b) {
    return Math.round((a + b)*100)/100;
  }, 0);

  // check if the partial sum is equals to target
  if (s === target) {
    console.log(originalList);
    partial.forEach(function(findIndex){
      console.log(findIndex, 'is at position ', originalList.indexOf(findIndex));
    });
  }

  for (var i = 0; i < numbers.length; i++) {
    n = numbers[i];
    remaining = numbers.slice(i + 1);
    // 递归调用时透传originalList
    subsetSum(remaining, target, partial.concat([n]), originalList);
  }
}

// 调用方式，不需要额外传原始数组，默认用第一个参数作为原始数组
subsetSum([3,9,8,4,5.5,7,10.07],15.57);

// 也可以主动指定原始数组，比如你要基于外部的其他数组获取索引
const myCustomList = [2,3,5,7,11];
subsetSum(myCustomList, 10, undefined, myCustomList);

方案2：用闭包封装避免递归传参 #

如果不想每次递归都透传参数，可以用外层函数把原始数组存在闭包里，代码更简洁：

function createSubsetSumSolver(originalList) {
  // 内部递归函数直接使用闭包中的originalList，不需要额外传参
  function subsetSum(numbers, target, partial) {
    var s, n, remaining;
    partial = partial || [];

    // sum partial
    s = partial.reduce(function (a, b) {
      return Math.round((a + b)*100)/100;
    }, 0);

    // check if the partial sum is equals to target
    if (s === target) {
      console.log(originalList);
      partial.forEach(function(findIndex){
        console.log(findIndex, 'is at position ', originalList.indexOf(findIndex));
      });
    }

    for (var i = 0; i < numbers.length; i++) {
      n = numbers[i];
      remaining = numbers.slice(i + 1);
      subsetSum(remaining, target, partial.concat([n]));
    }
  }

  // 返回调用入口，只需要传入目标和即可触发计算
  return function(target) {
    return subsetSum(originalList, target);
  }
}

// 调用方式
const solveSubsetSum = createSubsetSumSolver([3,9,8,4,5.5,7,10.07]);
solveSubsetSum(15.57);

注意事项 #

如果你的原始数组中存在重复数值，Array.indexOf()只会返回第一个匹配项的索引，会导致索引结果不准确。如果需要完全准确的原始索引，建议初始时把数值和索引绑定处理：

// 把原数组转成带索引的对象
const originalList = [3,9,8,4,5.5,7,10.07];
const indexedList = originalList.map((val, idx) => ({ value: val, index: idx }));

// 后续计算和的时候取value属性，匹配到结果后直接取index属性即可得到准确原始索引

内容的提问来源于stack exchange，提问作者lin ede