https://leetcode.cn/problems/maximum-score-of-spliced-array/description/
这一题应该算一个连续最大子数组思维题,要点是根据差数组去做,然后求最值
class Solution {
public:
int maximumsSplicedArray(vector<int>& nums1, vector<int>& nums2) {
// f[i]表示以a[i]为结尾的最大子数组和
// f[i]=max(f[i-1],0) + a[i];
// 根据nums1和nums2的差来求最大子数组和
const int N = 1e5+10;
int f[N];
int f2[N];
int nums[N];
memset(f,0,sizeof f);
memset(f2,0,sizeof f2);
for(int i=0;i<nums1.size();i++)nums[i]=nums1[i]-nums2[i];
int maxAns=0,minAns=0;
for(int i=1;i<=nums1.size();i++)
{
f[i]=max(f[i-1],0) + nums[i-1];
f2[i]=min(f2[i-1],0) + nums[i-1];
maxAns=max(f[i],maxAns);
minAns=min(f2[i],minAns);
printf("num=%d,maxAns=%d,minAns=%d\n",nums[i-1],maxAns,minAns);
}
int sum1=0,sum2=0;
for(int i=0;i<nums1.size();i++)
{
sum1+=nums1[i];
sum2+=nums2[i];
}
return max(abs(minAns)+sum1,maxAns+sum2);
}
};
标签:f2,int,memset,2321,拼接,数组,leetcode From: https://www.cnblogs.com/lxl-233/p/18134634