題目鏈接：https://leetcode.com/problems/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).
void Sort(int *data,int n) {<span style="white-space:pre"> </span>//歸併排序非遞歸實現
int *tmp = (int *)malloc(n * sizeof(int));
int lBegin, lEnd, rBegin, rEnd;
int i,j,k;
int len = 1;
while (len < n) {
lBegin = 0;
lEnd = len - 1;
rBegin = len;
while (rBegin < n) {
rEnd = lEnd + len < n - 1 ? lEnd + len : n - 1;
i = lBegin,j = rBegin,k = lBegin;
while (i <= lEnd && j <= rEnd) {
if (data[i] <= data[j])
tmp[k++] = data[i++];
else
tmp[k++] = data[j++];
}
while (i <= lEnd)
tmp[k++] = data[i++];
while (j <= rEnd)
tmp[k++] = data[j++];
for (i = lBegin; i <= rEnd; ++i)
data[i] = tmp[i];
lBegin += 2 * len;
lEnd += 2 * len;
rBegin += 2 * len;
}
len *= 2;
}
free(tmp);
}
int threeSumClosest(int* nums, int numsSize, int target) {
int i, j, k;
int sum, closestSum = nums[0] + nums[1] + nums[2];
Sort(nums, numsSize); //先排序
for(i = 0; i < numsSize; ++i) { //每趟一個數固定，另兩個數分別從剩下數列的第一個和最後一箇中間逼近
j = i + 1;
k = numsSize - 1;
while(j < k) {
sum = nums[i] + nums[j] + nums[k];
if(sum < target) { //和小於target，只有增大j才能使sum變大，從而更接近target
if(abs(sum-target) < abs(closestSum-target)) //如果更接近target更新closetSum
closestSum = sum;
++j;
}
else if(sum > target) { //和大於target，減小sum
if(abs(sum-target) < abs(closestSum-target))
closestSum = sum;
--k;
}
else //如果等於target是接近答案，直接返回結果
return sum;
}
}
return closestSum;
}