leetcode 3Sum. 處理特殊狀況+哈希思想

Given an array nums of n integers, are there elements a, b, c in nums
such that a + b + c = 0? Find all unique triplets in the array which
gives the sum of zero.

Note:

The solution set must not contain duplicate triplets.

Example:

Given array nums = [-1, 0, 1, 2, -1, -4],

A solution set is: [ [-1, 0, 1], [-1, -1, 2] ]

和TwoSum整體思想差很少，可是要多一層外循環，枚舉每個target的值，target的值爲0-nums[i], 咱們能夠作一些優化在減小遍歷次數：
1.正數直接跳出，由於排序後後面的確定都是正數
2.和前一個數相同的值能夠直接continue

注意幾種特殊狀況[-1,0,1,2,-1,-4]，排序後爲[-4,-1,-1,0,1,2],[-2,0,0,2,2],push的時候判斷下左指針和右指針與上個位置的是否相等，注意這裏的條件是或，若是有一個不相等，且值和爲target便可push

/**
 * @param {number[]} nums
 * @return {number[][]}
 */
var threeSum = function(nums) {
  let ans = [];
  nums.sort((a,b) => a-b || -1);
  let len = nums.length;
  for(let i = 0; i < len; i++) {
      if(nums[i] > 0)
          break;
      if(i>0 && nums[i]===nums[i-1])
          continue;
      let target = 0 - nums[i];
      let l = i+1,r=len-1;

      while(l<r) {
          if(nums[l]+nums[r]===target && (((l-1>=0)&&nums[l]!==nums[l-1])||(nums[r]!==nums[r+1]&&(r+1<=len))) ) {
              let t = [];
              t.push(nums[i],nums[l],nums[r]);
              ans.push(t);
              l++;
              r--;
          }else if(nums[l]+nums[r]<target) {
              l++;
          }else if(nums[l]+nums[r]>target) {
              r--;
          }else {
              l++;
              r--;
          }
      }
  }
  return ans;
};

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