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]
]
A solution set is:
[
[-1, 0, 0, 1],
[-2, -1, 1, 2],
[-2, 0, 0, 2]
]
求四个数的和的话可以降为求三个数的和
class Solution {
public:
vector<vector<int>> fourSum(vector<int>& nums, int target) {
vector<vector<int>> res;
if(nums.size()<4)
return res;
sort(nums.begin(),nums.end());
for(int i=0;i<nums.size()-1;i++){
cout<<"----------------"<<endl;
if(4*nums[i]>target) //值已经超了
break;
//int tar3=target-nums[i];
int tar3=target-nums[i];
//cout<<target<<" "<<i<<endl;
threeSum(nums,res,tar3,i);
}
return res;
}
void threeSum(vector<int>& nums,vector<vector<int>>&result,int tar3,int start){//start表示的是计算四个数的和的第一个数
if(start>0){
while(nums[start]==nums[start-1])
//如果这个和前面的相同的话，这个就直接结束好了，再继续下去反而会接着出现问题
return;
}
for(int i=start+1;i<nums.size()-1;i++){
cout<<i<<endl;
int target=tar3-nums[i];
int front=i+1;
int back=nums.size()-1;
while(front<back){
if(nums[front]+nums[back]==target){
vector<int> temp;
temp.push_back(nums[start]);
temp.push_back(nums[i]);
temp.push_back(nums[front]);
temp.push_back(nums[back]);
result.push_back(temp);
while(front<back&&nums[front]==temp[2]) front++;
while(front<back&&nums[back]==temp[3]) back--;
//哇塞哇塞在这里出现了死循环
//while(front<back&&nums[front]==temp[1]) front++;
//while(front<back&&nums[back]==temp[2]) back--;
}
else if(nums[front]+nums[back]<target)
front++;
else if(nums[front]+nums[back]>target)
back--;
}
while((nums[i]==nums[i+1])&&((i+1)<nums.size())){
i++;
}
}
}
};
求三个数的和仍然使用两个指针，但是由于加了一道步骤所以参数多了，再加上自己命名变量的时候名字有时候会发生重复，所以就导致有点懵。
参数命名的时候需要更加清晰一点啦
而且这次似乎有点得意忘形，所以写的马虎点，于是调了很久。
另外，bug还是要用纸笔一行一行的调。
这个代码是在讨论区看到的，他的优势是把各种情况都考虑的很全面，程序运行起来的话如果遇到这种情况就可以直接结束这次循环，而不是嵌套进去再一点一点的验证，但是这种频繁的判断语句是否也会浪费时间呢，两者相比起来哪个更快一点呢
呀，这个时候明白了计组和操作系统在代码编写中的作用，尤其是c，c++这样与底层相关的语言。
有一个问题，在使用容器的size时是每次都调用函数快还是在一开始就定义一个变量好呢，想必是第二种方式吧。
class Solution {
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;
}
}