3Sum

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

【Note】

1.Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c)

2.The solution set must not contain duplicate triplets.

For example, given array S = {-1 0 1 2 -1 -4},    A solution set is:    (-1, 0, 1)    (-1, -1, 2)

【代码】

public class Solution {    public static List<List<Integer>> threeSum(int[] nums) {        List<List<Integer>> result = new ArrayList<List<Integer>>();        if (nums.length<3)            return result;         sort(nums, 0, nums.length-1);       // Arrays.sort(nums);        List<Integer> temp = new ArrayList<Integer>();        if (nums[0] + nums[1] + nums[2] > 0)            return result;        if (nums[0] + nums[1] + nums[2] == 0) {            temp.add(nums[0]);            temp.add(nums[1]);            temp.add(nums[2]);            result.add(temp);            return result;        }        int begin = 0, mid, end;        while (begin < nums.length-1) {            while (begin > 0 && begin < nums.length -1&& nums[begin] == nums[begin - 1])                begin++;            mid = begin + 1;            end = nums.length - 1;            while (mid < end) {                if (nums[begin] + nums[mid] + nums[end] == 0) {                    List<Integer> temp1 = new ArrayList<Integer>();                    temp1.add(nums[begin]);                    temp1.add(nums[mid]);                    temp1.add(nums[end]);                    result.add(temp1);                    mid++;                    end--;                    while (mid < end && nums[mid] == nums[mid - 1])                        mid++;                    while (end > mid && nums[end] == nums[end + 1])                        end--;                } else if (nums[begin] + nums[mid] + nums[end] < 0) {                    mid++;                    while (mid < end && nums[mid] == nums[mid - 1])                        mid++;                }                else {                    end--;                    while (end > mid && nums[end] == nums[end + 1])                        end--;                }            }            begin++;        }        // HashSet h = new HashSet(result);        // result.clear();        // result.addAll(h);        return result;    }    public static int[] sort(int[] nums, int low, int high) {        int mid = (low + high) / 2;        if (low < high) {            sort(nums, low, mid);            sort(nums, mid + 1, high);            merge(nums, low, mid, high);        }        return nums;    }    public static void merge(int[] nums, int low, int mid, int high) {        int[] temp = new int[high - low + 1];        int i = low;// 左指针        int j = mid + 1;// 右指针        int k = 0;        while (i <= mid && j <= high) {            if (nums[i] < nums[j]) {                temp[k++] = nums[i++];            } else {                temp[k++] = nums[j++];            }        }        while (i <= mid) {            temp[k++] = nums[i++];        }        while (j <= high) {            temp[k++] = nums[j++];        }        for (int k2 = 0; k2 < temp.length; k2++) {            nums[k2 + low] = temp[k2];        }    }    }}

【算法思想】先对链表排序，也可以直接使用函数排序，再设前中后三个指针，前一个指针作为固定，中间指针最后一个指针从两端挪动，如果三个值和等于零，放入list里，如果小于零，中间指针后移，如果大于零，后面指针前移，直到找到所有的符合的值。

【总结】比较虐心的一道题，因为老是超时，后来上网查，发现必须将所有经过的相同的数据过滤掉，才能AC！