[Leetcode] 644. Maximum Average Subarray II 解题报告

题目：

Given an array consisting of 

n

 integers, find the contiguous subarray whose length
is greater than or equal to 

k

 that has the maximum average value. And you need
to output the maximum average value.

Example 1:

Input: [1,12,-5,-6,50,3], k = 4
Output: 12.75
Explanation:
when length is 5, maximum average value is 10.8,
when length is 6, maximum average value is 9.16667.
Thus return 12.75.

Note:

1 <= 

k

 <= 

n

 <=
10,000.
Elements of the given array will be in range [-10,000, 10,000].
The answer with the calculation error less than 10-5 will be accepted.
思路：

确实是一道很难的题目，我开始的时候一直想用DP解决，奈何死活找不出递推公式。后来参考了网上的二分查找思路，感觉还是挺巧妙的：

我们首先找出数组中的最小值left和最大值right，因为这将构成最大平均值的区间。然后每次找到left和right的平均值mid，对于mid，在数组内找是否存在一个长度大于等于k的连续区间，其平均值大于mid，如果存在，那么说明mid还不是最大平均值，所以修改左边界left；否则说明mid已经大于等于左右长度大于等于k的所有连续区间的最大平均值了，所以此时修改右边界right。这样当left和right逐渐收敛到一点的时候，该收敛点就是最大平均值。

该算法的空间复杂度是O(1)，时间复杂度是O(logm*n)，其中n是nums的长度，而m则是nums中最大值和最小值的差。虽然m有可能很大，但是被log了，所以该时间复杂度接近于O(n)，还是非常不错的。

代码：class Solution {
public:
double findMaxAverage(vector<int>& nums, int k) {
double left = INT_MAX, right = INT_MIN, mid;
for (int num : nums) {
right = max(right, double(num)); // the max value in the array
left = min(left, double(num)); // the min value in the array
}
while (left + 0.00001 < right) { //threshold
mid = left + (right - left) / 2;
if(islarger(nums, mid, k)) {
right = mid;
}
else {
left = mid;
}
}
return left;
}
private:
// Return true when mid is larger than or equal to the maximum average value;
// Return false when mid is smaller than the maximum average value.
bool islarger(vector<int>& nums, double mid, int k){
// sum: the sum from nums[0] to nums[i];
// prev_sum: the sum from nums[0] to nums[i-k];
// min_sum: the minimal sum from nums[0] to nums[j] ( 0=< j <= i-k );
double sum = 0, prev_sum = 0, min_sum = 0;
for(int i = 0; i < nums.size(); i++){
sum += nums[i] - mid;
if (i >= k) {
prev_sum += nums[i-k] - mid;
min_sum = min(prev_sum, min_sum);
}
if(i >= k-1 && sum > min_sum) { // there exists an range in which the average value is larger
return false;
}
}
return true;
}
};