Question

获得了java solution

public List<List<Integer>> threeSum(int[] num) {
    Arrays.sort(num);
    List<List<Integer>> res = new LinkedList<>(); 
    for (int i = 0; i < num.length-2; i++) {
        if (i == 0 || (i > 0 && num[i] != num[i-1])) {
            int lo = i+1, hi = num.length-1, sum = 0 - num[i];
            while (lo < hi) {
                if (num[lo] + num[hi] == sum) {
                    res.add(Arrays.asList(num[i], num[lo], num[hi]));
                    while (lo < hi && num[lo] == num[lo+1]) lo++;
                    while (lo < hi && num[hi] == num[hi-1]) hi--;
                    lo++; hi--;
                } else if (num[lo] + num[hi] < sum) lo++;
                else hi--;
           }
        }
    }
    return res;
}

我当前的javascript：

var mysort = function(a, b) {
  return a - b;
}

var threeSum = function(ns) {
  // everything is sorted
  ns.sort(mysort);

  // acc
  let res = [];

  // loop all #, but we keep last 2 elements
  for (let i = 0; i < ns.length - 2; i++) {

    // 1. i === 0, rm 1st element
    // 2. same same skip
    if (i === 0 || (i > 0 && ns[i] !== ns[i - 1])) {
      //if (true) {
      // the 2nd element
      let lo = i + 1;
      // the end element
      let hi = ns.length - 1;
      // remove the 1st element
      let sum = 0 - ns[i];

      // bi search
      while (lo < hi) {

        console.log(lo, hi, ns[lo], ns[hi], sum)

        // bi search: 2nd element + end element === sum
        if ((ns[lo] + ns[hi]) === sum) {
          console.log('push');
          res.push([ns[i], ns[lo], ns[hi]]);

          // skip: lo < hi, lo skip equal
          while (lo < hi && ns[lo] === ns[lo + 1]) lo++;

          // skip: lo < hi, hi skip equal
          while (lo < hi && ns[hi] === ns[hi - 1]) hi--;

          // closer
          lo++;

          // closer
          hi--;
        } else if (ns[lo] + ns[hi] < sum)
          lo++; // lo + hi < sum, lo++
        else
          hi--; // lo + hi > sum, hi--
      }
    }

    return res;
  }
}

console.log(threeSum([-1,0,1,2,-1,-4]));

无法通过第一个测试用例：

[-1,0,1,2,-1,-4]

Answer 1

您的代码很好，但您放错了return的位置：应该将其移出for循环。

所以：

    return res;
  }
}

...应该是：

  }
  return res;
}

请注意，在JavaScript中，您无需防止超出范围的数组引用，因为它只会返回undefined，它不等于任何数值。因此，您可以这样做：

if (ns[i] !== ns[i - 1])

即使在Java中，您也不需要执行此测试：i > 0 &&，因为受shortcut evaluation的影响，它在求值时总是正确的。因此，在Java版本中，您可以执行以下操作：

if (i == 0 || num[i] != num[i-1]) {

Answer 2

这是您问题的解决方案：

function threeSum(nums){

  let len = nums.length;
  if(len < 3) return [];
  nums.sort(function(a,b){
    return a-b;
  })
  if(nums[0] > 0 || nums[0] + nums[1] + nums[2] > 0) return [];
  if(len === 3) {
    if(nums[0] + nums[1] + nums[2] === 0) return [nums];
    else return [];
  }
  //console.log(nums);
  let result = [];
  let checker = '';
  for(let i=0; i<len-2; i++){
    for(let j=i+1; j<len-1; j++){
      //since the array is sorted, we only need to start from 
      //the last index that where value was found, since 
      //nums[k] has to be a lower number.
      for(let start = len-1, k = start; k>i, k>j;) {
        let triplet = [nums[i], nums[j], nums[k]];
        let tripletString = '/' + nums[i] + ' ' + nums[j] + ' ' + nums[k] + '/';
        if(nums[i] + nums[j] === -nums[k] && !checker.includes(tripletString)){
          result.push(triplet);
          checker += tripletString; 
          start--;
          k = start;
        } else {
          k--;
        }
      }
    }
  }
  return result;
}
var output = threeSum([-1,0,1,2,-1,-4]);
console.log(output);

Answer 3

如果删除类型定义，则JavaScript中的实现几乎相同

function threeSum(num) {
  // Arrays.sort(num);
  num.sort((a, b) => a-b);
  // List<List<Integer>> res = new LinkedList<>();
  const res = [];
  for (let i = 0; i < num.length-2; i++) {
      if (i == 0 || (i > 0 && num[i] != num[i-1])) {
          let lo = i+1, hi = num.length-1, sum = 0 - num[i];
          while (lo < hi) {
              if (num[lo] + num[hi] == sum) {
                  // push a new Array instead of .add List
                  res.push([num[i], num[lo], num[hi]]);
                  while (lo < hi && num[lo] == num[lo+1]) lo++;
                  while (lo < hi && num[hi] == num[hi-1]) hi--;
                  lo++; hi--;
              } else if (num[lo] + num[hi] < sum) lo++;
              else hi--;
         }
      }
  }
  return res;
}

尝试将Java中的3sum转换为javascript

3 个答案: