LeetCode problem solving within fifteen: Four and the number of

topic

Given a n array of integers and a target nums target, if there are four elements a, b, c, and d nums determined such that a + b + c + d is equal to the value of the target? Identify all satisfy the conditions of the quad and do not repeat.

Note: The answer can not contain duplicate quad.

Example:

给定数组 nums = [1, 0, -1, 0, -2, 2]，和 target = 0。

满足要求的四元组集合为：
[
  [-1,  0, 0, 1],
  [-2, -1, 1, 2],
  [-2,  0, 0, 2]
]

analysis

It is the general problem-solving ideas, before and after the two pointers while walking to get to the target value minus the value of the value again to continue to move backward or forward on the basis of that, to get added to the value of the previous step is 0, then in this case four values ​​are satisfactory.

answer

public class Solution {
    public List<List<Integer>> fourSum(int[] num, int target) {
		List<List<Integer>> rst = new ArrayList<List<Integer>>();
		Arrays.sort(num);

		for (int i = 0; i < num.length - 3; i++) {
			if (i != 0 && num[i] == num[i - 1]) {
				continue;
			}

			for (int j = i + 1; j < num.length - 2; j++) {
				if (j != i + 1 && num[j] == num[j - 1])
					continue;

				int left = j + 1;
				int right = num.length - 1;
				while (left < right) {
					int sum = num[i] + num[j] + num[left] + num[right];
					if (sum < target) {
						left++;
					} else if (sum > target) {
						right--;
					} else {
						ArrayList<Integer> tmp = new ArrayList<Integer>();
						tmp.add(num[i]);
						tmp.add(num[j]);
						tmp.add(num[left]);
						tmp.add(num[right]);
						rst.add(tmp);
						left++;
						right--;
						while (left < right && num[left] == num[left - 1]) {
							left++;
						}
						while (left < right && num[right] == num[right + 1]) {
							right--;
						}
					}
				}
			}
		}

		return rst;
	}
}