644. Maximum Average Subarray II

先看个简单的

Given an array consisting of n integers, find the contiguous subarray of given length 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 = 4Output: 12.75Explanation: Maximum average is (12-5-6+50)/4 = 51/4 = 12.75

Note:

1. 1 <= k <= n <= 30,000.
2. Elements of the given array will be in the range [-10,000, 10,000].

package l643;public class Solution {    public double findMaxAverage(int[] nums, int k) {        long sum = 0, max = 0;        for(int i=0; i<k; i++)sum+=nums[i];        max = sum;                for(int i=k; i<nums.length; i++) {        sum += nums[i];        sum -= nums[i-k];        if(sum > max)        max = sum;        }                return (max+0.0)/k;    }}

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 = 4Output: 12.75Explanation: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. 1 <= k <= n <= 10,000.
2. Elements of the given array will be in range [-10,000, 10,000].
3. The answer with the calculation error less than 10-5 will be accepted.

思路：有2个trick

1. 只要满足一定误差，所以用二分，直接二分最后的平均值

2. 在判断某个平均值是否满足时，转换为求连续子数组的和大于0（这个求平均值的套路可以记一下）

/* * 只要10-5 精度就好 */public class Solution {    public double findMaxAverage(int[] nums, int k) {    int min = Integer.MAX_VALUE, max = Integer.MIN_VALUE;    for(int i : nums) {    min = Math.min(min, i);    max = Math.max(max, i);    }        double lo = min, hi = max;    while(hi-lo > 1e-6) {    double mid = (lo+hi)/2.0;    if(ok(nums, k ,mid))    lo = mid;    else    hi = mid;    }        return lo;    }private boolean ok(int[] nums, int k, double mid) {// 数组每个数减去mid，转换为：连续子数组累加大于0double[] t = new double[nums.length];for(int i=0; i<nums.length; i++)t[i] = nums[i] - mid;double sum = 0;for(int i=0; i<k; i++)sum += t[i];if(sum >= 0)return true;double min = 0, sum2 = 0;for(int i=k; i<t.length; i++) {sum2 += t[i-k];sum += t[i];min = Math.min(min, sum2);if(sum - min >= 0)return true;}return false;}}