016 - 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).

给定一个数组，找出三个元素，相加的和最接近target

int mycompar(const void *a, const void *b)
{
int *x = (int *)a;
int *y = (int *)b;
if (*x > *y) return 1;
return *x == *y? 0 : -1;
}

int search(int *num, int size, int aim, int i1, int i2)
{
int left = 0, right = size - 1, mid;
while (left <= right) {
mid = (left + right) / 2;
if (num[mid] == aim) {
int x = mid, y = mid, i;
while (x >= 0 && num[x] == aim) x--;
while (y < size && num[y] == aim) y++;
for (i = x + 1; i < y; i++)
if (i != i1 && i != i2)
return aim;
break;
}
if (num[mid] > aim) right = mid - 1;
else left = mid + 1;
}
while (right >= 0) {
if (right == i1 || right == i2) right--;
else break;
}
while (left < size) {
if (left == i1 || left == i2) left++;
else break;
}
if (right >= 0 && left < size) {
return abs(num[left] - aim) > abs(num[right] - aim)? num[right] : num[left];
} else if(right < 0)
return num[left];
else
return num[right];
}

int threeSumClosest(int* nums, int numsSize, int target)
{
int close = INT_MAX / 2;
int i, j, k = 0;
qsort(nums, numsSize, sizeof(int), mycompar);

for (i = 0; i < numsSize - 2; i ++) {
while (i && i < numsSize - 2 && nums[i - 1] == nums[i]) i++;
if (i == numsSize - 2) break;
for (j = i + 1; j < numsSize - 1; j ++) {
while (j-1 > i && j < numsSize - 1 && nums[j] == nums[j - 1]) j++;
if (j == numsSize - 1) break;
k = search(nums, numsSize, target - nums[i] - nums[j], i, j);
k = nums[i] + nums[j] + k;
if (abs(target - k) < abs(close - target)) close = k;
if (close == target) return target;
}
}

return close;
}