643.Maximum Average Subarray I

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Solution 1: accepted 22ms

Iteration.
Time: O(n)
Space: O(1) 

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public double findMaxAverage(int[] nums, int k) {
    int len = nums.length;
    if (len < k) return 0;
    double sum = 0;
    for(int i = 0; i < k; i++) {
        sum += nums[i] / (double) k;
    }
    double max = sum;
    for(int i = 1; i < len - k + 1; i++) {
        sum = sum - nums[i-1] / (double) k + nums[i+k-1] / (double) k;
        max = Math.max(max, sum);
    }
    return max;  
}

Solution 2: accepted 19ms

Calculate the average at last - be flexible man. 

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public double findMaxAverage(int[] nums, int k) {
    int len = nums.length;
    if (len < k) return 0;
    double sum = 0;
    for(int i = 0; i < k; i++) {
        sum += nums[i];
    }
    double max = sum;
    for(int i = 1; i < len - k + 1; i++) {
        sum = sum - nums[i - 1] + nums[i + k - 1];
        max = Math.max(max, sum);
    }
    return max / (double) k;  
}