leetcode18~4Sum

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

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]
]

比求3Sum多一层嵌套循环，还有一种解法是使用hashmap，先缓存两个数的和，然后再求解。

下面给出了两种解法，第二种解法参考leetcode讨论区的代码，过滤条件比较多，最后运行时间真的很短，可以借鉴一下~

public class FourSum {

//时间复杂度O(n^3)
public List<List<Integer>> fourSum2(int[] nums, int target) {
List<List<Integer>> res = new ArrayList<>();
if(nums==null || nums.length<4) return res;
Arrays.sort(nums);
for(int i=0;i<nums.length-3;i++) {
if(i>0 && nums[i]==nums[i-1]) continue;
for(int j=i+1;j<nums.length-2;j++) {
if(j>i+1 && nums[j]==nums[j-1]) continue;
int top = j+1,tail= nums.length-1;
while(top<tail) {
if(nums[i]+nums[j]+nums[top]+nums[tail]==target) {
List<Integer> list = new ArrayList<Integer>();
list.add(nums[i]);
list.add(nums[j]);
list.add(nums[top]);
list.add(nums[tail]);
res.add(list);
while(top<tail && nums[top]==nums[top+1]) top++;
while(top<tail && nums[tail]==nums[tail-1]) tail--;
top++;
tail--;
} else if(nums[i]+nums[j]+nums[top]+nums[tail]<target) {
top++;
} else {
tail--;
}
}
}
}
return res;
}

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)
return res;

Arrays.sort(nums);

int max = nums[len - 1];
if (4 * nums[0] > target || 4 * max < 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;
}

/*
* Find all possible distinguished three numbers adding up to the target
* in sorted array nums[] between indices low and high. If there are,
* add all of them into the ArrayList fourSumList, using
* fourSumList.add(Arrays.asList(z1, the three numbers))
*/
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);
}

}

/*
* Find all possible distinguished two numbers adding up to the target
* in sorted array nums[] between indices low and high. If there are,
* add all of them into the ArrayList fourSumList, using
* fourSumList.add(Arrays.asList(z1, z2, the two numbers))
*/
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;
}
}