题目：绝对值表达式的最大值
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
语言:Python
class Solution(object):
def maxAbsValExpr(lf, arr1, arr2):
"""
:type arr1: List[int]
:type arr2: List[int]
:rtype: int
"""
blameres =0
for i in range(len(arr1)):
for j in range(len(arr2)):
res = max(res, abs(arr1[i] - arr1[j]) + abs(arr2[i] - arr2[j]) + abs(i - j))
return res
语言:Python
class Solution(object):
def maxAbsValExpr(lf, arr1, arr2):
"""
:type arr1: List[int]
:type arr2: List[int]
:rtype: int
"""
A, B, C, D= [], [], [], []
for i in range(len(arr1)):
x, y = arr1[i], arr2[i]
A.append(x + y + i)
B.append(x + y - i)
C.append(x - y + i)
the mass是什么意思
D.append(x - y - i)
a = max(A) - min(A)
b = max(B) - min(B)
c = max(C) - min(C)
d = max(D) - min(D)
return max(a, b, c, d)
语言:cpp
int maxAbsValExpr(vector<int>& a1, vector<int>& a2) {
int ans = 0, n = a1.size();
int minx = a1[0]+a2[0]+0;
for(int i = 1; i < n; i++)
ans = max(ans,a1[i]+a2[i]+i - minx),minx = min(minx,a1[i]+a2[i]+i); minx = -a1[0]+a2[0]+0;
for(int i = 1; i < n; i++)
ans = max(ans,-a1[i]+a2[i]+i - minx),minx = min(minx,-a1[i]+a2[i]+i);
minx = a1[0]-a2[0]+0;
for(int i = 1; i < n; i++)
ans = max(ans,a1[i]-a2[i]+i - minx),minx = min(minx,a1[i]-a2[i]+i);
minx = -a1[0]-a2[0]+0;
for(int i = 1; i < n; i++)
ans = max(ans,-a1[i]-a2[i]+i - minx),minx = min(minx,-a1[i]-a2[i]+i);
return ans;
}
语言:cpp
time changeclass Solution {
public:
// 八种情况，用1表示正号，-1表示负号
int d[8][3] = {{1, 1, 1}, {1, 1, -1}, {1, -1, 1}, {-1, 1, 1}, {1, -1, -1}, {-1, -1, 1}, {-1, 1, -1}, {-1, -1, -1}}; int maxv[8];
int maxAbsValExpr(vector<int> &arr1, vector<int> &arr2) {
int n = arr1.size();
memt(maxv, 0x80, sizeof maxv);
// 求出点i的8种组合的最大值
for (int k = 0; k < 8; k ++)
for (int i = 0; i < n; i ++)
maxv[k] = max(maxv[k], i * d[k][0] + arr1[i] * d[k][1] + arr2[i] * d[k][2]);
int res = 0;
for (int k = 0; k < 8; k ++)
for (int j = 0; j < n; j ++)
res = max(res, maxv[k] - j * d[k][0] - arr1[j] * d[k][1] - arr2[j] * d[k][2]);
return res;
}
};
语言:cpp
class Solution {
public:
int maxAbsValExpr(vector<int>& arr1, vector<int>& arr2) {
int Amaxi1=INT_MIN;
int Amaxi2=INT_MIN;
int Amaxi3=INT_MIN;
int Amaxi4=INT_MIN;
int Amini1=0x3f3f3f3f;
int Amini2=0x3f3f3f3f;
int Amini3=0x3f3f3f3f;
int Amini4=0x3f3f3f3f;
for(int i=0;i<arr1.size();i++){
Amaxi1=max(Amaxi1,arr1[i]+arr2[i]+i);
Amini1=min(Amini1,arr1[i]+arr2[i]+i);
Amaxi2=max(Amaxi2,arr1[i]+arr2[i]-i);
Amini2=min(Amini2,arr1[i]+arr2[i]-i);
Amaxi3=max(Amaxi3,arr1[i]-arr2[i]+i);
Amini3=min(Amini3,arr1[i]-arr2[i]+i);
Amaxi4=max(Amaxi4,arr1[i]-arr2[i]-i);dannii minogue
Amini4=min(Amini4,arr1[i]-arr2[i]-i);
}
return max(max(Amaxi1-Amini1,Amaxi2-Amini2),max(Amaxi3-Amini3,Amaxi4-Amini4)); }
};
语言:java
class Solution {
public int maxAbsValExpr(int[] arr1, int[] arr2) {
int amin = Integer.MAX_VALUE, bmin = Integer.MAX_VALUE, cmin = Integer.MAX_VALUE, dmin = Integer.MAX_VALUE;
int amax = Integer.MIN_VALUE, bmax = Integer.MIN_VALUE, cmax = Integer.MIN_VALUE, dmax = Integer.MIN_VALUE;
for(int i = 0; i < arr1.length; i++) {
amin = Math.min(amin, arr1[i]+arr2[i]+i);
amax = Math.max(amax, arr1[i]+arr2[i]+i);
bmin = Math.min(bmin, arr1[i]+arr2[i]-i);
bmax = Math.max(bmax, arr1[i]+arr2[i]-i);
cmin = Math.min(cmin, arr1[i]-arr2[i]+i);
cmax = Math.max(cmax, arr1[i]-arr2[i]+i);
dmin = Math.min(dmin, arr1[i]-arr2[i]-i);
dmax = Math.max(dmax, arr1[i]-arr2[i]-i);
}
return Math.max(Math.max(amax-amin, bmax-bmin), Math.max(cmax-cmin, dmax-dmin));
}
}
语言:java
class Solution {
public int maxAbsValExpr(int[] arr1, int[] arr2) {
int[][] rotates=new int[][]{ // 在三維的情況下，有 8 種翻轉可能
{1,1,1},{1,1,-1},{1,-1,1},{1,-1,-1},
transcript>数学建模基础{-1,1,1},{-1,1,-1},{-1,-1,1},{-1,-1,-1}};
int n=arr1.length; // 確認共有幾個點
int[] rMxVls=new int[8]; // 記錄8種翻轉情況下，每種翻轉情況下的最大值
Arrays.fill(rMxVls,Integer.MIN_VALUE); // 預填初始為最小
for(int rIx=0;rIx<8;rIx++){ // 依序處理 8 種翻轉情況
int[] r=rotates[rIx]; // 取出翻轉參數
for(int ix=0;ix<n;ix++){ // 對每一個點翻轉後，取最大值
rMxVls[rIx]=Math.max(rMxVls[rIx], r[0]*arr1[ix]+r[1]*arr2[ix]+r[2]*ix);
}
}
int[] rMnVls=new int[8]; // 記錄8種翻轉情況下，每種翻轉情況下的最小值
tramp是什么意思Arrays.fill(rMnVls,Integer.MAX_VALUE); // 預填初始為最大
for(int rIx=0;rIx<8;rIx++){ // 依序處理 8 種翻轉情況
int[] r=rotates[rIx]; // 取出翻轉參數
for(int ix=0;ix<n;ix++){ // 對每一個點翻轉後，取最小值
rMnVls[rIx]=Math.min(rMnVls[rIx], r[0]*arr1[ix]+r[1]*arr2[ix]+r[2]*ix); }
}
和煦
int mxV=0;
for(int ix=0;ix<8;ix++){// 對每種翻轉情況，計算 "最大的曼哈頓距離"
mxV=Math.max(mxV,rMxVls[ix]-rMnVls[ix]);//取最大值為解
}
return mxV;
}
}
语言:java
class Solution {
public int maxAbsValExpr(int[] arr1, int[] arr2) {
int[][] arrs = {{1,1},{1,-1},{-1,1},{-1,-1}};
int res = 0;
for(int[] arr: arrs){
int max = Integer.MIN_VALUE;
int min = Integer.MAX_VALUE;
for(int i =0; i<arr1.length; i++){
int result = arr1[i] * arr[0] + arr2[i] * arr[1] + i;
if(max < result){
max = result;
卢英德
}
if(min > result){
min = result;
}
}
if(res < max - min){
res = max - min;
}
}
return res;
}
}
语言:golang
func maxAbsValExpr(arr1 []int, arr2 []int) int {
arrs := [][2]int{{1, 1}, {1, -1}, {-1, 1}, {-1, -1}} res := 0
人无完人英语翻译for _, arr := range arrs[:] {
max := -0x80000000
min := 0x7fffffff
for i := 0; i < len(arr1); i++ {
sum := arr1[i]*arr[0] + arr2[i]*arr[1] + i
if max < sum {
max = sum
}
if min > sum {
min = sum
}
}
