16. 3Sum Closest
2017-09-26 23:36
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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.
本题与3Sum很类似,解出3Sum后再解此题,会觉得很简单。
程序如下:
class Solution {
public int threeSumClosest(int[] nums, int target) {
Arrays.sort(nums);
int retVal = 0, deltVal = Integer.MAX_VALUE;
for (int i = 2; i < nums.length; ++ i){
int twoSumTarget = target - nums[i];
int l = 0, r = i - 1, curDelt;
while (l < r){
curDelt = Math.abs(nums[l] + nums[r] - twoSumTarget);
if (deltVal > curDelt){
deltVal = curDelt;
retVal = nums[i] + nums[l] + nums[r];
}
if (nums[l] + nums[r] < twoSumTarget){
l ++;
}
else if (nums[l] + nums[r] > twoSumTarget){
r --;
}
else {
return target;
}
}
}
return retVal;
}
}
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).
本题与3Sum很类似,解出3Sum后再解此题,会觉得很简单。
程序如下:
class Solution {
public int threeSumClosest(int[] nums, int target) {
Arrays.sort(nums);
int retVal = 0, deltVal = Integer.MAX_VALUE;
for (int i = 2; i < nums.length; ++ i){
int twoSumTarget = target - nums[i];
int l = 0, r = i - 1, curDelt;
while (l < r){
curDelt = Math.abs(nums[l] + nums[r] - twoSumTarget);
if (deltVal > curDelt){
deltVal = curDelt;
retVal = nums[i] + nums[l] + nums[r];
}
if (nums[l] + nums[r] < twoSumTarget){
l ++;
}
else if (nums[l] + nums[r] > twoSumTarget){
r --;
}
else {
return target;
}
}
}
return retVal;
}
}
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