if(arr.empty()) return 0;
vector<double> maxv(arr.size()),minv(arr.size());
maxv[0]=minv[0]=arr[0];
double maxsum = arr[0];
for(int i = 1; i < arr.size(); i++) {
double ai = arr[i];
if(ai > 0) {
maxv[i] = max(ai, maxv[i-1]*ai);
minv[i] = min(ai, minv[i-1]*ai);
} else {
maxv[i] = max(ai, minv[i-1]*ai);
minv[i] = min(ai, maxv[i-1]*ai);
}
maxsum = max(maxsum, maxv[i]);
}
return maxsum;
发现只与前一个状态有关,可进行空间优化
double maxF = arr[0], minF = arr[0], ans = arr[0];
for (int i = 1; i < arr.size(); ++i) {
double mx = maxF, mn = minF;
maxF = max(mx * arr[i], max(arr[i], mn * arr[i]));
minF = min(mn * arr[i], min(arr[i], mx * arr[i]));
ans = max(maxF, ans);
}
return ans;