1131. 绝对值表达式的最大值

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最新推荐文章于 2024-07-14 20:59:54 发布

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分类专栏： leetcode C++

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本文介绍了一种算法，用于解决两个等长整数数组间的最大绝对值表达式问题。通过巧妙转换和优化，将原始问题转化为求解四个简化表达式的最大值，实现了高效求解。

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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

分析：

首先把绝对值符号去掉，展开表达式：

|arr1[i] - arr1[j]| + |arr2[i] - arr2[j]| + |i - 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)
 = -(arr1[i] - arr2[i] - i) + (arr1[j] - arr2[j] - j)

因为存在四组两两等价的展开，所以可以优化为四个表达式：

A = arr1[i] + arr2[i] + i
B = arr1[i] + arr2[i] - i
C = arr1[i] - arr2[i] + i
D = arr1[i] - arr2[i] - i

max( |arr1[i] - arr1[j]| + |arr2[i] - arr2[j]| + |i - j|)
= max(max(A) - min(A),
      max(B) - min(B),
      max(C) - min(C),
      max(D) - min(D))

代码实现：

class Solution {
public:
    int maxAbsValExpr(vector<int>& arr1, vector<int>& arr2) {
//         |arr1[i] - arr1[j]| + |arr2[i] - arr2[j]| + |i - j|
    /*
    A = arr1[i] + arr2[i] + i
    B = arr1[i] + arr2[i] - i
    C = arr1[i] - arr2[i] + i
    D = arr1[i] - arr2[i] - i

    max( |arr1[i] - arr1[j]| + |arr2[i] - arr2[j]| + |i - j|)
    = max(max(A) - min(A),
      max(B) - min(B),
      max(C) - min(C),
      max(D) - min(D))
    */
        int max_a,min_a,max_b,min_b,max_c,min_c,max_d,min_d;
        max_a=min_a=max_b=min_b=max_c=min_c=max_d=min_d=0;
        for(int i=0; i<arr1.size(); i++){
            int temp = arr1[i]+arr2[i]+i;
            max_a = temp>max_a?temp:max_a;
            min_a = temp<min_a?temp:min_a;
            
            temp = arr1[i]+arr2[i]-i;
            max_b = temp>max_b?temp:max_b;
            min_b = temp<min_b?temp:min_b;
            
            temp = arr1[i]-arr2[i]+i;
            max_c = temp>max_c?temp:max_c;
            min_c = temp<min_c?temp:min_c;
            
            temp = arr1[i]-arr2[i]-i;
            max_d = temp>max_d?temp:max_d;
            min_d = temp<min_d?temp:min_d;
        }
        int max = max_a - min_a;
        max = max_b-min_b>max?max_b-min_b:max;
        max = max_c-min_c>max?max_c-min_c:max;
        max = max_d-min_d>max?max_d-min_d:max;
        return max;
        
    }
};