1. 程式人生 > >18. 4Sum(求陣列中和為指定值的4個數)

18. 4Sum(求陣列中和為指定值的4個數)

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: 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]
]
題目大意:給定一個整型陣列和一個整數target,找到4個元素,和為target,返回這4個元素,要求返回的結果集不包含重複的方案。解題思路:參照《15. 3Sum(求陣列中和為0的3個數)》,先對陣列排序,不同的是按排序結果每次鎖定兩個數nums[i]和nums[j],然後用雙指標思想從這兩個數的右側找到兩個數,這兩個數的和為target-nums[i]-nums[j]。解題程式碼:(67ms,beats 61.09%)
class Solution {
    public List<List<Integer>> fourSum(int[] nums, int target) {
        List<List<Integer>> res = new ArrayList<>();
		Arrays.sort(nums);
		int len = nums.length;
		for (int i = 0; i < len - 3; i++) {
			for (int j = i + 1; j < len - 2; j++) {
				int left = j + 1;
				int right = len - 1;
				int diff = target - nums[i] - nums[j];
				while (left < right) {
					if (nums[left] + nums[right] == diff) {
						res.add(new ArrayList<>(Arrays.asList(nums[i], nums[j], nums[left], nums[right])));
						left++;
						while (left < len && nums[left] == nums[left - 1])
							left++;
						right--;
						while (right > left && nums[right] == nums[right + 1])
							right--;
					} else if (nums[left] + nums[right] < diff) {
						left++;
					} else {
						right--;
					}
				}
				while (j < len - 3 && nums[j] == nums[j + 1]) {
					j++;
				}
			}
			while (i < len - 4 && nums[i] == nums[i + 1]) {
				i++;
			}
		}
		return res;
    }
}