Given an arraynumsofn_integers and an integertarget, are there elements_a,b,c, andd_innumssuch that_a+b+c+d=target? Find all unique quadruplets in the array which gives the sum oftarget.
Note: The solution set must not contain duplicate quadruplets.
Example
Given array nums = [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]
]
Note
主要是小优化问题
每一轮的指针都要注意去重;
每一轮进入内循环之前,可以直接看一眼 index 与
index 后面的连续几个数
nums[] 末尾的最后几个数
是不是会比 target 小或者大,如果 index 与后面连续的几个数加起来都没 target 大,那么检查这个 index 的必要都没有,可以直接跳过;如果 index 与最末尾的几个数相加依然小于 target,那么说明我们应该直接去看下一个更大的数当 index.
Code
publicclassSolution {publicList<List<Integer>> fourSum(int[] nums,int target) {List<List<Integer>> total =newArrayList<>();int n =nums.length;if(n <4) return total;Arrays.sort(nums);for (int i =0; i < n -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[n -3] + nums[n -2] + nums[n -1] < target) continue;for (int j = i +1; j < n -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[n -2] + nums[n -1] < target) continue;int left = j +1, right = n -1;while (left < right) {int sum = nums[left] + nums[right] + nums[i] + nums[j];if (sum < target) left++;elseif (sum > target) right--;else {total.add(Arrays.asList(nums[i],nums[j],nums[left++],nums[right--]));while(left < right && nums[left] == nums[left -1]) left ++;while(left < right && nums[right] == nums[right +1]) right --; } } } }return total; }}