题目描述:
示例 1:
- 将第二个元素替换为第一个元素:[1,7,5] => [1,1,5] ,或者
- 将第二个元素替换为第三个元素:[1,7,5] => [1,5,5]
两种方案的绝对差值和都是 |1-2| + (|1-3| 或者 |5-3|) + |5-5| = 3
输入:nums1 = [2,4,6,8,10], nums2 = [2,4,6,8,10]
输出:0
解释:nums1 和 nums2 相等,所以不用替换元素。绝对差值和为 0
示例 3:
题目分析:
- 要求最小绝对差值之和
- 只可以替换nums1中的一个数
- nums1中替换的元素,与
思路一:
代码实现:
class Solution {
public int minAbsoluteSumDiff(int[] nums1, int[] nums2) {
final int mod = 1000000007;
int len = nums1.length;
int minNum = 0, index = 0, mark = 0;
// 求nums1中的替换的那个元素
for (int i = 0; i < len; i++) {
for (int j = 0; j < len; j++) {
int num = Math.abs(nums2[i] - nums1[j]);
int val = Math.abs(nums2[i] - nums1[i]);
// 只有差值小于val才有替换的需要
if (num >= val) {
continue;
}
// 判断替换的值是否是最小绝对差值
int result = val - num;
if (minNum < result) {
minNum = result;
index = j;
mark = i;
}
}
}
// 替换
nums1[mark] = nums1[index];
// 求最小绝对差值之和
int sum = 0;
for (int i = 0; i < len; i++) {
sum += Math.abs(nums1[i] - nums2[i]);
}
return sum % mod;
}
}
思路二:
代码实现:
class Solution {
public int minAbsoluteSumDiff(int[] nums1, int[] nums2) {
final int mod = 1000000007;
int len = nums1.length;
int[] arr = new int[len];
System.arraycopy(nums1, 0, arr, 0, len);
Arrays.sort(arr);
int sum = 0, maxNum = 0;
for (int i = 0; i < len; i++) {
int diff = Math.abs(nums1[i] - nums2[i]);
sum = (sum + diff) % mod;
int index = binary(arr, nums2[i]);
maxNum = Math.max(maxNum, diff - index);
}
return (sum - maxNum + mod) % mod;
}
public int binary(int[] arr, int target) {
int low = 0, high = arr.length - 1;
if (arr[high] <= target) {
return target - arr[high];
} else if (arr[low] >= target) {
return arr[low] - target;
}
while (low < high) {
int mid = (high - low) / 2 + low;
if (arr[mid] < target) {
low = mid + 1;
} else {
high = mid;
}
}
int minNum = Math.min(target - arr[low - 1], arr[low] - target);
return minNum;
}
}