Given an array nums of n integers and an integer target, are there elements a, b, c, and d in nums 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.
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] ]
code1：（暴力一点的）
List<List<Integer>> result=new ArrayList<>();
public List<List<Integer>> fourSum(int[] nums, int target) {
if(nums.length<4)
return result;
Arrays.sort(nums);
for(int i=0;i<=nums.length-4;){
for(int j=i+1;j<=nums.length-3;){
int front=j+1;
int last=nums.length-1;
while(front<last){
int sum=nums[front]+nums[last]+nums[i]+nums[j];
if(sum==target){
List<Integer> list=new ArrayList<>();
list.add(nums[i]);
list.add(nums[j]);
list.add(nums[front]);
list.add(nums[last]);
result.add(list);
front++;
while(front<last&&nums[front]==nums[front-1])
front++;
last--;
while(front<last&&nums[last]==nums[last+1])
last--;
}else if(sum<target){
front++;
while(front<last&&nums[front]==nums[front-1])
front++;
}else{
last--;
while(front<last&&nums[last]==nums[last+1])
last--;
}
}
j++;
while(j<nums.length&&nums[j]==nums[j-1])
j++;
}
i++;
while(i<nums.length&&nums[i]==nums[i-1])
i++;
}
return result;
}code2（优化后的）
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)
while(i<j&&nums[++i]==nums[i-1])
;
if (sum > target)
while(i<j&&nums[--j]==nums[j+1])
;
}
return;
}
code3：（优化后的代码短的）
public List<List<Integer>> fourSum(int[] num, int target) {
ArrayList<List<Integer>> ans = new ArrayList<>();
if(num.length<4)return ans;
Arrays.sort(num);
for(int i=0; i<num.length-3; i++){
if(num[i]+num[i+1]+num[i+2]+num[i+3]>target)break; //first candidate too large, search finished
if(num[i]+num[num.length-1]+num[num.length-2]+num[num.length-3]<target)continue; //first candidate too small
if(i>0&&num[i]==num[i-1])continue; //prevents duplicate result in ans list
for(int j=i+1; j<num.length-2; j++){
if(num[i]+num[j]+num[j+1]+num[j+2]>target)break; //second candidate too large
if(num[i]+num[j]+num[num.length-1]+num[num.length-2]<target)continue; //second candidate too small
if(j>i+1&&num[j]==num[j-1])continue; //prevents duplicate results in ans list
int low=j+1, high=num.length-1;
while(low<high){
int sum=num[i]+num[j]+num[low]+num[high];
if(sum==target){
ans.add(Arrays.asList(num[i], num[j], num[low], num[high]));
while(low<high&&num[low]==num[low+1])low++; //skipping over duplicate on low
while(low<high&&num[high]==num[high-1])high--; //skipping over duplicate on high
low++;
high--;
}
//move window
else if(sum<target)
while(low<high&&num[++low]==num[low-1])
;
else
while(low<high&&num[--high]==num[high+1])
;
}
}
}
return ans;
}