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  • [leetcode 18]4Sum

    1 题目:

    Given an array S of n integers, are there elements abc, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

    Note:

    • Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
    • The solution set must not contain duplicate quadruplets.
        For example, given array S = {1 0 -1 0 -2 2}, and target = 0.
    
        A solution set is:
        (-1,  0, 0, 1)
        (-2, -1, 1, 2)
        (-2,  0, 0, 2)

    2 思路

    首先,得理解3Sum。

    然后按照3Sum的思路,就好做了。主要是要遍历所有的四个组合。排序是肯定的。详情见代码吧,自己想的。。时间复杂度o(n^3)做了几题,感觉有点状态了。

    3 代码

        public List<List<Integer>> fourSum(int[] nums, int target) {
            List<List<Integer>> result = new ArrayList<List<Integer>>();
            int len = nums.length;
            if(len < 4){
                return result;
            }
            
            Arrays.sort(nums);
            int head = 0;
            int end = len-1;
            while(head < end -2){
                while(head < end-2){
                    int pre = head + 1;
                    int suffix = end - 1;
                    /*  compare all four group between pre & suffic   */
                    while(pre < suffix){
                        int sum = nums[head]+nums[end]+nums[pre]+nums[suffix];
                        if(sum < target){
                            ++pre;
                        }else if(sum > target){
                            --suffix;
                        }else{
                            result.add(Arrays.asList(nums[head],nums[pre],nums[suffix],nums[end]));
                            ++pre;
                            --suffix;
                            while(pre <= suffix && nums[pre]==nums[pre-1]) ++pre;
                            while(pre <= suffix && nums[suffix]==nums[suffix+1]) --suffix;
                        }
                    }
                    /* for each nums[head], compore all nums[head] until head<end-2 */
                    ++head;
                    while(head<end-2 && nums[head]==nums[head-1]) ++head;
                }
                /* set head to 0 & end-1 ,so we can traverse all four groups */
                head=0;
                --end;
                while(head<end-2 &&nums[end] == nums[end+1]) --end;
            }
        
            return result;
        }
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  • 原文地址:https://www.cnblogs.com/lingtingvfengsheng/p/4779531.html
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