【双指针】LeetCode 18. 四数之和

题目链接

18. 四数之和

思路

【双指针】LeetCode 15. 三数之和 思路相似,但是要注意测试用例可能溢出,需要转换为 long

代码

class Solution {
    public List<List<Integer>> fourSum(int[] nums, int target) {
        Arrays.sort(nums);
        List<List<Integer>> result = new ArrayList<>();

        for(int a = 0; a < nums.length - 3; a++){
            if(a > 0 && nums[a] == nums[a - 1]){
                continue;
            }
            for(int b = a + 1; b < nums.length - 2; b++){
                if(b > a + 1 && nums[b] == nums[b - 1]){
                    continue;
                }

                int c = b + 1;
                int d = nums.length - 1;
                while(c < d){
                    long sum = (long)nums[a] + (long)nums[b] + (long)nums[c] + (long)nums[d];
                    if(sum == target){
                        result.add(Arrays.asList(nums[a], nums[b], nums[c], nums[d]));
                        while(c < d && nums[c + 1] == nums[c]){
                            c++;
                        }
                        while(c < d && nums[d - 1] == nums[d]){
                            d--;
                        }
                        c++;
                        d--;
                    }else if(sum > target){
                        d--;
                    }else{
                        c++;
                    }
                }

            }
        }

        return result;
    }
}
posted @ 2023-01-30 09:56  Frodo1124  阅读(19)  评论(0编辑  收藏  举报