LeetCode No18. 四数之和

题目

给你一个由 n 个整数组成的数组 nums ，和一个目标值 target 。请你找出并返回满足下述全部条件且不重复的四元组 [nums[a], nums[b], nums[c], nums[d]] （若两个四元组元素一一对应，则认为两个四元组重复）：

0 <= a, b, c, d < n
a、b、c 和 d 互不相同
nums[a] + nums[b] + nums[c] + nums[d] == target
你可以按 任意顺序 返回答案 。

示例 1：

输入：nums = [1,0,-1,0,-2,2], target = 0
输出：[[-2,-1,1,2],[-2,0,0,2],[-1,0,0,1]]

示例 2：

输入：nums = [2,2,2,2,2], target = 8
输出：[[2,2,2,2]]

提示：

1 <= nums.length <= 200
-10^9 <= nums[i] <= 10^9
-10^9 <= target <= 10^9

思路

暴力

四个for循环，排除下重复数据即可。

双指针

思路参考：LeetCode No15.三数之和

AC代码

暴力

点击查看代码

class Solution {
    public List<List<Integer>> fourSum(int[] nums, int target) {
        List<List<Integer>> res = new ArrayList<>();
        List<Integer> list = new ArrayList<>();
        int n = nums.length;
        Arrays.sort(nums);
        int lasta = 100000000+1; 
        for(int i=0; i<n; i++) {
            int a = nums[i];
            if( a==lasta ) {
                lasta = a;
                continue;
            }
            lasta = a;
            int lastb = 100000000+1;
            for(int j=i+1; j<n; j++) {
                int b = nums[j];
                if( b==lastb ) {
                    lastb = b;
                    continue;
                }
                lastb = b;
                int lastc = 100000000+1;
                for(int k=j+1; k<n; k++) {
                    int c = nums[k];
                    if( lastc==c ) {
                        lastc = c;
                        continue;
                    }
                    lastc = c;
                    int lastd = 100000000+1;
                    for(int l=k+1; l<n; l++) {
                        int d = nums[l];
                        if( lastd!=d && a+b+c+d == target ) {
                            list = new ArrayList<>();
                            list.add(a);
                            list.add(b);
                            list.add(c);
                            list.add(d);
                            res.add(list);
                            break;
                        }
                        lastd = d;
                    }
                }
            }
        }
        return res;
    }
}

### 双指针

点击查看代码

class Solution {
    public List<List<Integer>> fourSum(int[] nums, int target) {
        List<List<Integer>> quadruplets = new ArrayList<List<Integer>>();
        if (nums == null || nums.length < 4) {
            return quadruplets;
        }
        Arrays.sort(nums);
        int length = nums.length;
        for (int i = 0; i < length - 3; i++) {
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            if ((long) nums[i] + nums[i + 1] + nums[i + 2] + nums[i + 3] > target) {
                break;
            }
            if ((long) nums[i] + nums[length - 3] + nums[length - 2] + nums[length - 1] < target) {
                continue;
            }
            for (int j = i + 1; j < length - 2; j++) {
                if (j > i + 1 && nums[j] == nums[j - 1]) {
                    continue;
                }
                if ((long) nums[i] + nums[j] + nums[j + 1] + nums[j + 2] > target) {
                    break;
                }
                if ((long) nums[i] + nums[j] + nums[length - 2] + nums[length - 1] < target) {
                    continue;
                }
                int left = j + 1, right = length - 1;
                while (left < right) {
                    int sum = nums[i] + nums[j] + nums[left] + nums[right];
                    if (sum == target) {
                        quadruplets.add(Arrays.asList(nums[i], nums[j], nums[left], nums[right]));
                        while (left < right && nums[left] == nums[left + 1]) {
                            left++;
                        }
                        left++;
                        while (left < right && nums[right] == nums[right - 1]) {
                            right--;
                        }
                        right--;
                    } else if (sum < target) {
                        left++;
                    } else {
                        right--;
                    }
                }
            }
        }
        return quadruplets;
    }
}