LeetCode 1131. 绝对值表达式的最大值（数学绝对值展开）

文章目录

1. 题目
2. 解题

1. 题目

给你两个长度相等的整数数组，返回下面表达式的最大值：

|arr1[i] - arr1[j]| + |arr2[i] - arr2[j]| + |i - j|

其中下标 i，j 满足 0 <= i, j < arr1.length。

示例 1：
输入：arr1 = [1,2,3,4], arr2 = [-1,4,5,6]
输出：13示例 2：
输入：arr1 = [1,-2,-5,0,10], arr2 = [0,-2,-1,-7,-4]
输出：20提示：
2 <= arr1.length == arr2.length <= 40000
-10^6 <= arr1[i], arr2[i] <= 10^6

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/maximum-of-absolute-value-expression
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2. 解题

只有这8种可能
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;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;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;

把 j 拿走
arr1[i]+arr2[i]+i;−arr1[i]−arr2[i]−i;−arr1[i]+arr2[i]+i;arr1[i]−arr2[i]+i;arr1[i]+arr2[i]−i;−arr1[i]−arr2[i]+i;arr1[i]−arr2[i]−i;−arr1[i]+arr2[i]−i;arr1[i] + arr2[i] + i;\\ -arr1[i] - arr2[i] - i;\\ -arr1[i] + arr2[i] + i;\\ arr1[i] - arr2[i] + i;\\ arr1[i] + arr2[i] - i;\\ -arr1[i] - arr2[i] + i;\\ arr1[i] - arr2[i] - i;\\ -arr1[i] + arr2[i] - i;\\ arr1[i]+arr2[i]+i;−arr1[i]−arr2[i]−i;−arr1[i]+arr2[i]+i;arr1[i]−arr2[i]+i;arr1[i]+arr2[i]−i;−arr1[i]−arr2[i]+i;arr1[i]−arr2[i]−i;−arr1[i]+arr2[i]−i;
发现只有4种情况
arr1[i]+arr2[i]+iasA;−arr1[i]−arr2[i]−ias−A;−arr1[i]+arr2[i]+iasB;arr1[i]−arr2[i]+iasC;arr1[i]+arr2[i]−iasD;−arr1[i]−arr2[i]+ias−D;arr1[i]−arr2[i]−ias−B;−arr1[i]+arr2[i]−ias−C;arr1[i] + arr2[i] + i \quad as A;\\ -arr1[i] - arr2[i] - i \quad as -A;\\ -arr1[i] + arr2[i] + i \quad as B;\\ arr1[i] - arr2[i] + i \quad as C;\\ arr1[i] + arr2[i] - i \quad as D;\\ -arr1[i] - arr2[i] + i \quad as -D;\\ arr1[i] - arr2[i] - i \quad as -B;\\ -arr1[i] + arr2[i] - i \quad as -C; arr1[i]+arr2[i]+iasA;−arr1[i]−arr2[i]−ias−A;−arr1[i]+arr2[i]+iasB;arr1[i]−arr2[i]+iasC;arr1[i]+arr2[i]−iasD;−arr1[i]−arr2[i]+ias−D;arr1[i]−arr2[i]−ias−B;−arr1[i]+arr2[i]−ias−C;
只需要取出ABCD，求其最大最小值，做差

class Solution {public:int maxAbsValExpr(vector<int>& arr1, vector<int>& arr2) {int min1 = INT_MAX, min2 = INT_MAX, min3 = INT_MAX, min4 = INT_MAX;int max1 = INT_MIN, max2 = INT_MIN, max3 = INT_MIN, max4 = INT_MIN;for(int i = 0; i < arr1.size(); ++i){max1 = max(max1, arr1[i] + arr2[i] + i);max2 = max(max2,-arr1[i] + arr2[i] + i);max3 = max(max3, arr1[i] - arr2[i] + i);max4 = max(max4, arr1[i] + arr2[i] - i);min1 = min(min1, arr1[i] + arr2[i] + i);min2 = min(min2,-arr1[i] + arr2[i] + i);min3 = min(min3, arr1[i] - arr2[i] + i);min4 = min(min4, arr1[i] + arr2[i] - i);}return max(max(max1-min1, max2-min2),max(max3-min3, max4-min4));}
};

100 ms 23.2 MB

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