public List<List<Integer>> fourSum(int[] nums, int target) {
Arrays.sort(nums);
//System.out.println(Arrays.toString(nums));
int len = nums.length;
List<List<Integer>> result = new LinkedList<>();
for(int i=0;i<len-3;i++){
if(i==0||nums[i]!=nums[i-1]){ //去除重复值
for(int j=i+1;j<len-2;j++){
if(j==i+1||nums[j]!=nums[j-1]){
int low = j+1,high=len-1;
while(low<high){
int temp = nums[i]+nums[j]+nums[low]+nums[high];
if(temp==target){
result.add(Arrays.asList(nums[i],nums[j],nums[low],nums[high]));
while(low<high&&nums[low]==nums[low+1]) low++;
while(low<high&&nums[high]==nums[high-1]) high--;
low++;
high--;
}
else if(temp<target){
low++;
}
else{
high--;
}
}
}
}
}
}
return result;
类似这种k-sum的题,都简化为sorted array的2sum问题,即利用two pointers, 一个头,一个尾,互相逼近看是否达到target值。
3-sum问题即,先选定一个,再在剩余数组中做2-sum,4-sum同上。注意代码中注释的如何去除重复值。
类似题目
2-sum:
public int[] twoSum(int[] nums, int target) {
Map<Integer, Integer> map = new HashMap<>();
int[] res = new int[2];
for(int i=0,len=nums.length;i<len;i++){
if(map.containsKey(target-nums[i])){
res[1]=i;
res[0]=map.get(target-nums[i]);
return res;
}
map.put(nums[i],i);
}
return res;
}
3-sum:
public List<List<Integer>> threeSum(int[] nums) {
List<List<Integer>> result = new LinkedList<>();
Arrays.sort(nums);
for(int i=0,len=nums.length;i<len-2;i++){
if(i==0||nums[i]!=nums[i-1]){
int low =i+1,high=nums.length-1;
while(low<high){
int temp = nums[i]+nums[low]+nums[high];
if(temp==0){
result.add(Arrays.asList(nums[i],nums[low],nums[high]));
while(low<high&&nums[low]==nums[low+1]) low++;
while(low<high&&nums[high]==nums[high-1]) high--;
low++;
high--;
}
else if(temp>0){
high--;
}
else low++;
}
}
}
return result;
}
3-sum closer:
Arrays.sort(nums);
int cur = 0,dis=Integer.MAX_VALUE;
for(int i=0,len=nums.length;i<len-2;i++){
if(i==0||nums[i]!=nums[i-1]){
int low = i+1,high=nums.length-1;
while(low<high){
int temp = nums[i]+nums[low]+nums[high];
int d = Math.abs(temp-target);
//cur = d<dis?temp:cur;
if(temp==target) return target;
else if(temp>target){
high--;
}
else{
low++;
}
//System.out.println(d+" "+dis);
if(d<dis){
dis=d;
cur=temp;
}
//cur = d<dis?temp:cur;
//System.out.println(cur);
}
}
}
return cur;
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