标签：nbsp double return false solution search log ret size
LintCode
给出一个整数数组，有正有负。找到这样一个子数组，他的长度大于等于 k，且平均值最大。
二分平均值,平均值的上限r是单个最大值,下限l是所有数的和.
mid=(l+r)/2
设sum[i]=nums[0]+nums[1]+nums[2]...nums[i]-i*mid;
保存m=min(sum[0],sum[1],sum[2]...sum[i-k+1].
如果发现sum[i]>m,说明在num[i-k+1]到num[i]之间的数,平均值大于mid.所以sum才能上涨.
class Solution {
public:
double search(vector<int>&nums, int k, double mid) {
vector<double> sum(nums.size() + 1);
double min = 0;
for (int i = 1; i <= nums.size(); i++) {
sum[i] = sum[i - 1] + (double)nums[i - 1] - mid;
if (i >= k && sum[i] >= min)
return true;
if (i >= k)
min = ::min(min, sum[i - k + 1]);
}
return false;
}
double maxAverage(vector<int>& nums, int k) {
vector<int> sum(nums.size());
sum[0] = nums[0];
for (int i = 1; i < nums.size(); i++)
sum[i] = sum[i - 1] + nums[i];
double l = (double)sum.back() / nums.size(), r = *max_element(nums.begin(), nums.end());
while (r - l > 1e-6) {
double mid = (l + r) / 2.0;
if (search(nums, k, mid))
l = mid;
else
r = mid;
}
return r;
}
};
标签：nbsp double return false solution search log ret size
原文地址：http://www.cnblogs.com/yujinding/p/7054757.html
