3Sum Closest

Notes:

跟3Sum差不多；

1、对数组排序；

2、对于每个a = num[i]，问题是在num[i+1]到num[n-1]中 b+c  = -a:

记b = num[i + 1]，c = num[n-1]；

sum = a + b + c;

若sum < target, and sum < small, samll = sum,  j++;

若sum > target,and sum > big, big = sum,  k--;

若sum == target，则返回target;

public class Solution {
        public int threeSumClosest(int[] nums, int target) {
          
          int a, b, c, sum;
          int i, j, k;
          int small = -2147483648, medium = -2147483648, big = 2147483647;
          int INT_MAX = 2147483647, INT_MIN = -2147483648;
          Arrays.sort(nums);
//          boolean flag = false;
          for (i = 0; i < nums.length; i++) {
//        	  flag = false;
          
        	 if(i == 0 || nums[i] != nums[i-1]){
        			a = nums[i];
        			j = i + 1;
        			k = nums.length - 1;
        			while(j < k){
        				b = nums[j];
        				c = nums[k];
        				
        				sum = a + b + c;
        	        	 if(sum < target){
        	        		 if(sum > small){
        	        			 small = sum;
        	        		 }
        	        		 j++;
        	        	 }
        	        	 else if(sum == target){
        	        		 return target;
        	        	 }
        	        	 else {
        	        		 if(sum < big){
        	        			 big = sum;
        	        		 }
        	        		 
        					k--;
        				}
        			}
        	 }
          }
          
          
          if(small != INT_MIN && big == INT_MAX){
        	  return small;
          }
          if(small == INT_MIN && big != INT_MAX){
        	  return big;
          }
          
          return (target - small)<(big - target)?small:big;
      }
}