15. 3Sum [Medium]

3Sum

Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.

Notice that the solution set must not contain duplicate triplets.

Constraints:

• 3 <= nums.length <= 3000
• -10^5 <= nums[i] <= 10^5

Example

Input: nums = [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]
Explanation: 
nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0.
nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.
nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.
The distinct triplets are [-1,0,1] and [-1,-1,2].
Notice that the order of the output and the order of the triplets does not matter.

思路

分析下题干

• 一组结果中元素不能重复使用
• 不能出现相同结果集

那应该需要想到一定按照顺序来，才能保证结果集完整，同时也能忽略相同的值，因为排完序，他们一定是相邻的

	[-4,-1,-1,0,1,2]
	-4 -> [-1,-1,0,1,2]  取出一个值作为定值，那现在这个问题就变成了Two Sum了
	...
	-4 -1 -1 0 [1,2]     依次遍历

题解

• 无脑快速AC

  public List<List<Integer>> threeSum(int[] nums) {
        Arrays.sort(nums);
        ArrayList<List<Integer>> result = new ArrayList<>();
        int left, right, cur;

        for (int i = 0; i < nums.length; i++) {
            cur = nums[i];
            left = i + 1;
            right = nums.length - 1;
	    // Two Sum II 解法
            while (left < right) {
                int sum = cur + nums[left] + nums[right];
                if (sum > 0)
                    right--;
                else if (sum < 0)
                    left++;
                else {
                    result.add(Arrays.asList(cur, nums[left], nums[right]));
                    left++;
		    // 如果当前left和下一位相等，那就应该跳过下一位，不然下一位也会被当作结果加进来
                    while (left < right) {
                        if (nums[left] == nums[left - 1])
                            left++;
                        else break;
                    }
                }
            }
	    // 如果当前定值和下一位相等，也需要跳过，不然结果集会相同
            while (i < nums.length - 1) {
                if (nums[i] == nums[i + 1])
                    i++;
                else break;
            }
        }
        return result;
    }

• 优化

   public List<List<Integer>> threeSum1(int[] nums) {
        Arrays.sort(nums);
        ArrayList<List<Integer>> result = new ArrayList<>();
        int left, right, cur;

        for (int i = 0; i < nums.length; i++) {
            cur = nums[i];
            if (i != 0 && cur == nums[i - 1])
                continue;
            left = i + 1;
            right = nums.length - 1;
            while (left < right) {
                int sum = cur + nums[left] + nums[right];
                if (sum > 0)
                    right--;
                else if (sum < 0)
                    left++;
                else {
                    result.add(Arrays.asList(cur, nums[left], nums[right]));
                    left++;
                    while (nums[left] == nums[left - 1] && left < right)
                        left++;
                }
            }
        }
        return result;
    }