一,题目
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).
二,思路
方法一:枚举+剪枝
方法二:two pointer
方法二比方法一了一层循环
三,代码
public int threeSumClosest(int[] nums, int target) {
if (nums == null || nums.length < 3) {
return 0;
}
int len = nums.length;
int diff = Integer.MAX_VALUE;
int sum = 0;
Arrays.sort(nums);
for (int i = 0; i < len - 2; i++) {
for (int j = i + 1; j < len - 1; j++) {
int right = j+1;
while (right < len) {
int tempSum = nums[i] + nums[j] + nums[right];
int tempDiff = tempSum - target;
int abs = Math.abs(tempDiff);
if (abs < diff) {
sum = tempSum;
diff = abs;
}
if(tempDiff > 0 && tempDiff > diff){
break;
}
right++;
}
}
}
return sum;
}
public int threeSumClosest2(int[] nums, int target){
if (nums == null || nums.length < 3) {
return 0;
}
Arrays.sort(nums);
int len = nums.length;
int bestSum = nums[0]+nums[1]+nums[2];
for(int i=0; i<len-2 ;i++){
int left = i+1;
int right = len -1;
while(left < right){
int sum = nums[i]+nums[left]+nums[right];
if(Math.abs(sum - target) < Math.abs(bestSum-target)){
bestSum = sum;
}
if(sum > target){
right --;
}else{
left++;
}
}
}
return bestSum;
}