leetcode 1131. 绝对值表达式的最大值

题意：

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。

思路：

根据数学思想，对于绝对值符号内的表达式( i > j )，共有四种可能

（1）arr1[i] > arr1[j] , arr2[i] > arr2[j]：arr1[i] - arr1[j] + arr2[i] - arr2[j] + i - j = ( arr1[i] + arr2[i] + i ) - ( arr1[j] + arr2[j] + j )

（2）arr1[i] > arr1[j] , arr2[i] < arr2[j]：arr1[i] - arr1[j] - arr2[i] + arr2[j] + i - j = ( arr1[i] - arr2[i] + i ) - ( arr1[j] - arr2[j] + j )

（3）arr1[i] < arr1[j] , arr2[i] > arr2[j]：- arr1[i] + arr1[j] + arr2[i] - arr2[j] + i - j = ( - arr1[i] + arr2[i] + i ) - ( - arr1[j] + arr2[j] + j )

（4）arr1[i] < arr1[j] , arr2[i] < arr2[j]：- arr1[i] + arr1[j] - arr2[i] + arr2[j] + i - j = ( - arr1[i] - arr2[i] + i ) - ( - arr1[j] - arr2[j] + j )

对于每一组（i，j），表达式的值必然属于上面四种情况的一种，并且所得到的值是四种情况中最大的，因此我们可以扫一遍数组，用当前值减去之前最小的值，再取一个最大值，就是所求的答案。

代码：

const int N=4e4+10;
class Solution {
public:
    int v[N][4],pre[4];
    int maxAbsValExpr(vector<int>& arr1, vector<int>& arr2) {
        int n = arr1.size();
        int ans=INT_MIN;
        for(int i = 0;i < n;i++){
            if(i==0){
                pre[0]=arr1[i] + arr2[i] + i;
                pre[1]=arr1[i] - arr2[i] + i;
                pre[2]=-arr1[i] + arr2[i] + i;
                pre[3]=-arr1[i] - arr2[i] + i;
            }
            else {
                ans=max(ans,arr1[i] + arr2[i] + i - pre[0]); pre[0]=min(pre[0],arr1[i] + arr2[i] + i);
                ans=max(ans,arr1[i] - arr2[i] + i - pre[1]); pre[1]=min(pre[1],arr1[i] - arr2[i] + i);
                ans=max(ans,-arr1[i] + arr2[i] + i - pre[2]); pre[2]=min(pre[2],-arr1[i] + arr2[i] + i);
                ans=max(ans,-arr1[i] - arr2[i] + i-pre[3]); pre[3]=min(pre[3],-arr1[i] - arr2[i] + i);
            }
        }
        return ans;
    }
};