3Sum Closest

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).

思想：将3Sum简化为2Sum，采用从前和从后同时查找的方式进行搜索；
注意事项：优化，先排序后，对相邻相同数字的优化；其次对向前和向后相同数字的优化。
JAVA 代码:

1. public int threeSumClosest(int[] num, int target) {
2. int len = num.length;
3. if(len<3) return 0;
4. // float min = (float)Integer.MAX_VALUE;
5. int min = num[0] + num[1] + num[2];  //注意min的选取
6. Arrays.sort(num);
7. for(int i=0;i<=len-3;i++) {
8. if(i>0 && num[i] == num[i-1]) continue;
9. int start = i+1;
10. int end = len-1;
11. while(start<end){
12. if(num[start] + num[end] + num[i] == target) {
13. return target;
14. } else {
15. if(Math.abs(num[start] + num[end] + num[i] - target) < Math.abs(min-target)) {
16. min = num[start] + num[end] + num[i];
17. }
18. if(num[start] + num[end] + num[i] > target) {
19. end--;
20. while(end>=0 && num[end]==num[end+1]) end--;
21. } else {
22. start++;
23. while(start<len&&num[start]==num[start-1]) start++;
24. }
25. }
26. }
27. }
28. return min;
29. }

未优化的C++代码：

1. int threeSumClosest(vector<int> &num, int target) {
2. if(num.size()<3) return 0;
3. sort(num.begin(),num.end());
4. int sz = num.size();
5. int minSum = num[0]+num[1]+num[2];
6. for(int i=0;i<sz-2;i++) {
7. int tmp=num[i];
8. int start = i+1;
9. int end=sz-1;
10. while(start<end) {
11. int sum=tmp+num[start]+num[end];
12. if(sum==target) return sum;
13. if(abs(target-sum)<abs(minSum-target)) minSum= sum;
14. if(sum>target) {
15. end--;
16. } else {
17. start++;
18. }
19. }
20. }
21. return minSum;
22. }