LeetCode – 3Sum Closest (Java)
Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.
For example, given array S = {-1 2 1 -4}, and target = 1. The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
O(n^2)
Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.
For example, given array S = {-1 2 1 -4}, and target = 1. The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
O(n^2)
public class Solution { public int threeSumClosest(int[] num, int target) { int min = Integer.MAX_VALUE; int result = 0; Arrays.sort(num); for (int i = 0; i < num.length; i++) { int j = i + 1; int k = num.length - 1; while (j < k) { int sum = num[i] + num[j] + num[k]; int diff = Math.abs(sum - target); if (diff < min) { min = diff; result = sum; } if (sum <= target) { j++; } else { k--; } } } return result; } }https://xuezhashuati.blogspot.com/2017/05/16-3sum-closest.html
Sort the array and use two pointers to find the sum of three numbers.
We maintain a closest number to target. Each time if we find the sum is closer, we update the closest number.
Time complexity is O(n2).
public int threeSumClosest(int[] nums, int target) { if (nums == null || nums.length < 3) { return -1; } Arrays.sort(nums); int closest = nums[0] + nums[1] + nums[2]; for (int i = 0; i < nums.length - 2; i++) { int left = i + 1; int right = nums.length - 1; while (left < right) { int sum = nums[i] + nums[left] + nums[right]; if (Math.abs(sum - target) < Math.abs(closest - target)) { closest = sum; } if (sum < target) { left++; } else { right--; } } } return closest; }http://n00tc0d3r.blogspot.com/2013/01/2sum-3sum-4sum-and-variances.html
https://leetcode.com/discuss/16335/java-solution-with-o-n2-for-reference
https://discuss.leetcode.com/topic/1978/a-n-2-solution-can-we-do-better
Similar to 3 Sum problem, use 3 pointers to point current element, next element and the last element. If the sum is less than target, it means we have to add a larger element so next element move to the next. If the sum is greater, it means we have to add a smaller element so last element move to the second last element. Keep doing this until the end. Each time compare the difference between sum and target, if it is less than minimum difference so far, then replace result with it, otherwise keep iterating.
public int threeSumClosest(int[] num, int target) {
int result = num[0] + num[1] + num[num.length - 1];
Arrays.sort(num);
for (int i = 0; i < num.length - 2; i++) {
int start = i + 1, end = num.length - 1;
while (start < end) {
int sum = num[i] + num[start] + num[end];
if (sum > target) {
end--;
} else {
start++;
}
if (Math.abs(sum - target) < Math.abs(result - target)) {
result = sum;
}
}
}
return result;
}public int threeSumClosest(int[] nums, int target) {
if (num.length <= 3) { for (int i: num) { sum += i; } return sum; } Arrays.sort(nums); int sum = nums[0] + nums[1] + nums[nums.length - 1]; int closestSum = sum; for(int i = 0; i < nums.length - 2; i++){ if(i==0 || nums[i]!=nums[i-1]){ int left = i + 1, right = nums.length - 1; while(left < right){ sum = nums[left] + nums[right] + nums[i]; if(sum < target){ //move closer to target sum. while(left<right && nums[left] == nums[left+1]){ left++; } left++; }else if(sum > target){ //move closer to target sum. while(left<right && nums[right] == nums[right-1]){ right--; } right--; }else{ return sum; } //update the closest sum if needed. if(Math.abs(target - sum) < Math.abs(target - closestSum)){ closestSum = sum; } } } } return closestSum; }
http://www.cnblogs.com/springfor/p/3860175.html
1 public int threeSumClosest(int[] num, int target) {
2 if(num==null || num.length<3)
3 return 0;
4
5 int min = Integer.MAX_VALUE;
6 int val = 0;
7 Arrays.sort(num);
8 for(int i = 0; i<=num.length-3;i++){
9 int low = i+1;
10 int high = num.length-1;
11 while(low<high){
12 int sum = num[i]+num[low]+num[high];
13 if(Math.abs(target-sum)<min){
14 min = Math.abs(target-sum);
15 val = sum;
16 }
18 if(target==sum){
19 return val;
20 }else if(target>sum){
21 low++;
22 }else{
23 high--;
24 }
25 }
26 }
27 return val;
28 }
http://bangbingsyb.blogspot.com/2014/11/leetcode-3sum-closest.html
int threeSumClosest(vector<int> &num, int target) { if(num.size()<3) return INT_MAX; sort(num.begin(),num.end()); int minDiff = INT_MAX; for(int i=0; i<num.size()-2; i++) { int left=i+1, right = num.size()-1; while(left<right) { int diff = num[i]+num[left]+num[right]-target; if(abs(diff)<abs(minDiff)) minDiff = diff; if(diff==0) break; else if(diff<0) left++; else right--; } } return target+minDiff; }http://rleetcode.blogspot.com/2014/02/given-array-s-n-find-three-integers-in.html
public int threeSumClosest(int[] num, int target) {if (num==null||num.length==0){return -1;}int result=0;int minDif=Integer.MAX_VALUE;Arrays.sort(num);for (int i=0; i<num.length-2; i++ ){int first=num[i];int left=i+1;int right=num.length-1;while (left<right){int value=first+num[left]+num[right];if (value==target){return value;}int diff=Math.abs(value-target);if (diff<minDif){result=value;minDif=diff;}if (value>target){do {right--;} while(left<right && num[right]==num[right+1]);}else if (value<target){do {left++;} while (left<right && num[left]==num[left-1]);}}}return result;}
Just another inefficient way
http://n00tc0d3r.blogspot.com/2013/01/2sum-3sum-4sum-and-variances.html
1: public int threeSumClosest(int[] num, int target) {
2: if (num.length<3) return 0;
3: Arrays.sort(num);
4: int res = num[0] + num[1] + num[2];
5: for (int i=0; i<num.length-2; ++i) {
6: // skip duplicates
7: if (i>0 && num[i]==num[i-1]) continue;
8: for (int j=i+1; j<num.length-1; ++j) {
9: // skip duplicates
10: if (j>i+1 && num[j]==num[j-1]) continue;
11: // binary search for the third element
12: int start=j+1, end=num.length-1;
13: while (start<=end) {
14: int mid = (start+end)/2;
15: int sum = num[i] + num[j] + num[mid];
16: int diff = sum - target;
17: if ( Math.abs(diff) < Math.abs(res - target) ) {
18: res = sum;
19: }
20: if (diff == 0) { // find the target
21: return res;
22: } else if (diff < 0) {
23: start = mid + 1;
24: } else {
25: end = mid - 1;
26: }
27: }
28: }
29: }
30: return res;
31: } 