Given two arrays of integers with equal lengths, return the maximum value of:
|arr1[i] - arr1[j]| + |arr2[i] - arr2[j]| + |i - j|
where the maximum is taken over all 0 <= i, j < arr1.length.
Example 1:
Input: arr1 = [1,2,3,4], arr2 = [-1,4,5,6]
Output: 13
Example 2:
Input: arr1 = [1,-2,-5,0,10], arr2 = [0,-2,-1,-7,-4]
Output: 20
Constraints:
2 <= arr1.length == arr2.length <= 40000-10^6 <= arr1[i], arr2[i] <= 10^6
这道题说是给了两个长度相等的数组 arr1 和 arr2，让找出这个式子 |arr1[i] - arr1[j]| + |arr2[i] - arr2[j]| + |i - j| 的最大值。仔细观察这个式子，实际上是要找出两个坐标i和j，然后分别在两个数组中找出对应位置的数字，求差的绝对值并相加。存在绝对值的话就不太好求极值，需要去掉，那么就可能有正负两种情况，每个绝对值都有两种情况，这里三个绝对值号总共就有八种情况：
arr1[i] - arr1[j] + arr2[i] - arr2[j] + i - j
arr1[i] - arr1[j] + arr2[i] - arr2[j] - i + j
arr1[i] - arr1[j] - arr2[i] + arr2[j] + i - j
arr1[i] - arr1[j] - arr2[i] + arr2[j] - i + j
- arr1[i] + arr1[j] + arr2[i] - arr2[j] + i - j
- arr1[i] + arr1[j] + arr2[i] - arr2[j] - i + j
- arr1[i] + arr1[j] - arr2[i] + arr2[j] + i - j
- arr1[i] + arr1[j] - arr2[i] + arr2[j] - i + j
合并一下，就是：
(arr1[i] + arr2[i] + i) - (arr1[j] + arr2[j] + j)
(arr1[i] + arr2[i] - i) - (arr1[j] + arr2[j] - j)
(arr1[i] - arr2[i] + i) - (arr1[j] - arr2[j] + j)
(arr1[i] - arr2[i] - i) - (arr1[j] - arr2[j] - j)
- (arr1[i] - arr2[i] - i) + (arr1[j] - arr2[j] - j)
- (arr1[i] - arr2[i] + i) + (arr1[j] - arr2[j] + j)
- (arr1[i] + arr2[i] - i) + (arr1[j] + arr2[j] - j)
- (arr1[i] + arr2[i] + i) + (arr1[j] + arr2[j] + j)
仔细观察上面八种情况，其实后四种和前四种是重复的，因为i和j是可以交换的。这样的话我们只要找出前四种情况的最大值即可，为了最大化，就需要尽可能的最大化前面括号中的值，最小化后面括号的值。好就好在两个括号中的值的计算方法是相同的。若将括号内的整体看作某个数组中的下标为i的数字的话，那么就是数组中的最大值减去最小值。现在就要构建这四种不同的数组，构建方法也非常直接，就是根据括号中的内容，从 arr1 和 arr2 中各取一个i位置的数字，相加或相减，然后加上或减去坐标i即可，放到四个不同的数组中，然后分别算出它们的最大值减去最小值，取其中的最大值返回即可，参见代码如下：
class Solution {
public:
int maxAbsValExpr(vector<int>& arr1, vector<int>& arr2) {
int n = arr1.size();
vector<int> sum1(n), sum2(n), diff1(n), diff2(n);
for (int i = 0; i < n; ++i) {
sum1[i] = arr1[i] + arr2[i] + i;
sum2[i] = arr1[i] + arr2[i] - i;
diff1[i] = arr1[i] - arr2[i] + i;
diff2[i] = arr1[i] - arr2[i] - i;
}
return max(max(helper(sum1), helper(sum2)), max(helper(diff1), helper(diff2)));
}
int helper(vector<int>& arr) {
int mx = arr[0], mn = arr[0];
for (int num : arr) {
mx = max(mx, num);
mn = min(mn, num);
}
return mx - mn;
}
};
Github 同步地址:
https://github.com/grandyang/leetcode/issues/1131
参考资料：
https://leetcode.com/problems/maximum-of-absolute-value-expression/