就是对于每个b_i来说都去找离它最近的那个a_k，然后gain_i就是|b_i - a_i| - |b_i - a_k|，意为如果替换当前的b对应的a能达到的最大收益。 找到最大的gain，sum(diff)减去其即可。我用了二分去找离b_i最近的a_k，不清楚如果直接找会不会超时，不妨试试。
public int findminAbs(int[] a, int[] b) { int res = 0; //for every bi, we need to find the cloasest in a //so we need to use binary search for sorted array a, that closest in break; Arrays.sort(a); for (int num: b) { int ak = binarySearch(num, a); res += Math.abs(num - ak); } return res; } private int binarySearch(int num, int[] a) { if (num < a[0]) { return a[0]; } if (num > a[a.length - 1]) { return a[a.length - 1]; } int lo = 0; int hi = a.length - 1; while (lo <= hi) { int mid = (hi - lo) / 2 + lo; if (value < a[mid]) { hi = mid - 1; } else if (value > a[mid]) { lo = mid + 1; } else { return a[mid]; } } //the final position of lo and hi will be lo == hi + 1 return (a[lo] - value) < (value - a[hi]) ? a[lo] : a[hi]; }