[LeetCode] #18 四数之和

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]] ：

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 。

与[LeetCode] #15 三数之和相似

需要再套一层循环

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

优化

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 (nums[i] + nums[i + 1] + nums[i + 2] + nums[i + 3] > target) {
                break;
            }
            if (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 (nums[i] + nums[j] + nums[j + 1] + nums[j + 2] > target) {
                    break;
                }
                if (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;
    }
}

知识点：无

总结：无

标签：四数,target,nums,int,18,length,right,LeetCode,left
来源： https://www.cnblogs.com/jpppp/p/15387828.html