整体思路同之前的一样,依然采取降低维度的方式进行
public List<List<Integer>> solution(int nums[], int target) {
List<List<Integer>> result = new ArrayList<>();
if((nums.length<4)||(nums==null))
{
return result;
}
Arrays.sort(nums);
if ((4*nums[0]>target)||(4*nums[nums.length-1]<target))
{
return result;
}//上面两个队特例的快速结束一定要加
int N = nums.length;
for (int i = 0; i < N; i++) {
int[] temnum = new int[N - 1];
System.arraycopy(nums, 0, temnum, 0, i);
System.arraycopy(nums, i + 1, temnum, i, N - i - 1);
result = threeSum(temnum, target - nums[i], nums[i], result);
}
return result;
} public List<List<Integer>> threeSum(int[] nums, int target, int num,List<List<Integer>> result) {
int sum;
for (int i = 0; i < nums.length - 2; i++) {
int start = i + 1, end = nums.length - 1;
while (start < end) {
sum = nums[i] + nums[start] + nums[end];
if (sum < target) {
start++;
continue;
}
if (sum > target) {
end--;
continue;
}
//下面的部分为对每一结果内部进行进行排序,这样在去重时方便
if (sum == target) {
List<Integer> list = new ArrayList<>();
if(num<nums[i])
list.add(num);
list.add(nums[i]);
if(num>=nums[i] && num<nums[start])
list.add(num);
list.add(nums[start]);
if ( num>=nums[start] &&num<nums[end])
list.add(num);
list.add(nums[end]);
if (num>=nums[end])
list.add(num);
start++;
end--;
if (result.contains(list))
continue;
result.add(list);
}
}
}
return result;
}
整体速度在leetcode上大概为50%