3. Two Pointers

好题：

• 3 sum

3 Sum

• two pointer去重的经典题

• 想一遍过也不容易，多练

• use 3 pointers, i, j, k

• i前向

• j, k对撞

• i from 0 ~n - 2， j from i+1 ~ k ，k from n - 1 to j

• if (i != 0 && numbers[i] == numbers[i - 1]) { continue; }

• 注意 edge case [0,0,0,0]

public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> res = new ArrayList<>();
        if (nums == null || nums.length < 3) return res;
        Arrays.sort(nums);
        int n = nums.length;
        for (int i = 0; i < n - 2; i++) {
            if (i != 0 && nums[i] == nums[i-1]) continue;
            int target = -nums[i];
            int j = i + 1, k = n - 1;
            while (j < k) {
                int sum = nums[j] + nums[k];
                if (sum == target) {
                    List<Integer> list = new ArrayList<>();
                    list.add(nums[i]);
                    list.add(nums[j]);
                    list.add(nums[k]);
                    res.add(list);
                    //remove duplicate
                    while (j + 1 < k && nums[j] == nums[j+1]) {
                        j++;
                    }
                    while (j < k - 1 && nums[k-1] == nums[k]) {
                        k--;
                    }
                    j++;
                    k--;
                } else if (sum < target) {
                    j++;
                } else {
                    k--;
                }
            }
        }
        return res;
    }