LeetCode 18 4Sum (4个数字之和等于target)

题目链接 https://leetcode.com/problems/4sum/?tab=Description

找到数组中满足 a+b+c+d=0的所有组合，要求不重复。

Basic idea is using subfunctions for 3sum and 2sum, and keeping throwing all impossible cases. O(n^3) time complexity, O(1) extra space complexity.

首先进行判断，由于是四个数相加等于target，其中可以通过判断剪去一些不必要的分支：

for (i = 0; i < len; i++) {
z = nums[i];
if (i > 0 && z == nums[i - 1])// avoid duplicate
continue;
if (z + 3 * max < target) // z is too small
continue;
if (4 * z > target) // z is too large
break;
if (4 * z == target) { // z is the boundary
if (i + 3 < len && nums[i + 3] == z)
res.add(Arrays.asList(z, z, z, z));
break;
}

threeSumForFourSum(nums, target - z, i + 1, len - 1, res, z);
}

进入3个数相加的同时，需要对target进行更新。

for (i = low; i < high - 1; i++) {
z = nums[i];
if (i > low && z == nums[i - 1]) // avoid duplicate
continue;
if (z + 2 * max < target) // z is too small
continue;

if (3 * z > target) // z is too large
break;

if (3 * z == target) { // z is the boundary
if (i + 1 < high && nums[i + 2] == z)
fourSumList.add(Arrays.asList(z1, z, z, z));
break;
}

twoSumForFourSum(nums, target - z, i + 1, high, fourSumList, z1, z);
}

进入两个数相加时，问题得到进一步简化。

int i = low, j = high, sum, x;
while (i < j) {
sum = nums[i] + nums[j];
if (sum == target) {
fourSumList.add(Arrays.asList(z1, z2, nums[i], nums[j]));

x = nums[i];
while (++i < j && x == nums[i]) // avoid duplicate
;
x = nums[j];
while (i < --j && x == nums[j]) // avoid duplicate
;
}
if (sum < target)
i++;
if (sum > target)
j--;
}

参考代码：

package leetcode_50;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

/***
*
* @author pengfei_zheng
* 四个数加法等于target
*/
public class Solution18 {
public List<List<Integer>> fourSum(int[] nums, int target) {
ArrayList<List<Integer>> res = new ArrayList<List<Integer>>();
int len = nums.length;
if (nums == null || len < 4) //不足4个元素
return res;

Arrays.sort(nums);//排序

int max = nums[len - 1];
if (4 * nums[0] > target || 4 * max < target)//4个最小值之和超过target或者4个最大值之和小于target
return res;

int i, z;
for (i = 0; i < len; i++) {
z = nums[i];
if (i > 0 && z == nums[i - 1])// avoid duplicate
continue;
if (z + 3 * max < target) // z is too small
continue;
if (4 * z > target) // z is too large
break;
if (4 * z == target) { // z is the boundary
if (i + 3 < len && nums[i + 3] == z)
res.add(Arrays.asList(z, z, z, z));
break;
}

threeSumForFourSum(nums, target - z, i + 1, len - 1, res, z);
}

return res;
}

public void threeSumForFourSum(int[] nums, int target, int low, int high, ArrayList<List<Integer>> fourSumList,
int z1) {
if (low + 1 >= high)
return;

int max = nums[high];
if (3 * nums[low] > target || 3 * max < target)
return;

int i, z;
for (i = low; i < high - 1; i++) {
z = nums[i];
if (i > low && z == nums[i - 1]) // avoid duplicate
continue;
if (z + 2 * max < target) // z is too small
continue;

if (3 * z > target) // z is too large
break;

if (3 * z == target) { // z is the boundary
if (i + 1 < high && nums[i + 2] == z)
fourSumList.add(Arrays.asList(z1, z, z, z));
break;
}

twoSumForFourSum(nums, target - z, i + 1, high, fourSumList, z1, z);
}

}

public void twoSumForFourSum(int[] nums, int target, int low, int high, ArrayList<List<Integer>> fourSumList,
int z1, int z2) {

if (low >= high)
return;

if (2 * nums[low] > target || 2 * nums[high] < target)
return;

int i = low, j = high, sum, x;
while (i < j) {
sum = nums[i] + nums[j];
if (sum == target) {
fourSumList.add(Arrays.asList(z1, z2, nums[i], nums[j]));

x = nums[i];
while (++i < j && x == nums[i]) // avoid duplicate
;
x = nums[j];
while (i < --j && x == nums[j]) // avoid duplicate
;
}
if (sum < target)
i++;
if (sum > target)
j--;
}
return;
}
}

Full Code