请用算法实现，从给定的无序、不重复的数组data中，取出n个数，使其相加和为sum。并给出算法的时间/空间复杂度。(不需要找到所有的解，找到一个解即可)

[lgwebdream]

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[lgwebdream]

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[jsm1003]

经典回溯法
function getAllCombin(array, n, sum) {
let res = null;

function isValid(item, track) {
return track.indexOf(item) === -1
}

function backtrack(nums, track, sum1) {
if(track.length === n && sum1 === sum) {
res = track.slice();
return;
}

for(let i = 0; i< nums.length; i++) {
  if(isValid(nums[i], track)) {
    track.push(nums[i])
    backtrack(nums, track, sum1 + nums[i])
    track.pop();
  }
}

}

backtrack(array, [], 0)
return res;
}

[GolderBrother]

function getResult(data, n, sum, temp = []) {
    if (temp.length === n) {
        const isExist = temp.reduce((accu, cur) => accu + cur, 0) === sum;
        if (isExist) return temp;
        return false;
    }
    let result = [];
    for (let i = 0, len = data.length; i < len; i++) {
        const current = data.shift();
        temp.push(current);
        result = getResult(data, n, sum, temp);
        if (result) break;
        // 还原为原数组
        temp.pop();
        data.push(current);
    }
    return result;
}
const arr = [1, 5, 6, 2, 4, 7, 3];
console.log(getResult(arr, 3, 10, []));

[GuoLizhi]

回溯算法

    // n数之和
    function nSum(arr = [], target, n) {
      const set = new Set()
      let result = null

      function backTack(currArr, currSum, currIndex) {
        // 回溯终止条件
        if (currSum === target && currArr.length === n) {
          result = [...currArr]
          return
        }
        for (let i = currIndex; i < arr.length; i++) {
          // 如果这个数之前已经使用过了，那直接跳过
          if (set.has(arr[i])) continue
          currArr.push(arr[i])
          set.add(arr[i])
          backTack(currArr, currSum + arr[i], currIndex + 1)
          // 回溯
          currArr.pop()
          set.delete(arr[i])
        }
        // 没有找到结果
        return null
      }

      backTack([], 0, 0)

      return result
    }