Maximum Subarray IV

最新推荐文章于 2024-11-10 17:43:48 发布

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最新推荐文章于 2024-11-10 17:43:48 发布

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本文介绍了一种算法，用于寻找整数数组中长度大于等于k的最大连续子数组和，并提供了Java和Python实现。如果数组元素少于k，则返回0。

Given an integer arrays, find a contiguous subarray which has the largest sum and length should be greater or equal to given length k.
Return the largest sum, return 0 if there are fewer than k elements in the array.

Notice

Ensure that the result is an integer type.

Have you met this question in a real interview?

Yes

Example

Given the array [-2,2,-3,4,-1,2,1,-5,3] and k = 5, the contiguous subarray [2,-3,4,-1,2,1] has the largest sum = 5.

java

public class Solution {
    /*
     * @param nums: an array of integer
     * @param k: an integer
     * @return: the largest sum
     */
    public int maxSubarray4(int[] nums, int k) {
        // write your code here
        if (nums == null || nums.length == 0 || k <= 0 || nums.length < k) {
            return 0;
        }
        int[] preSum = new int[nums.length + 1];
        preSum[0] = 0;
        int max = Integer.MIN_VALUE;
        int min = 0;
        for (int i = 1; i <= nums.length; i++) {
            preSum[i] = preSum[i - 1] + nums[i - 1];
            if (i >= k && max < preSum[i] - min) {
                max = preSum[i] - min;
            }
            if (i >= k) {
                min = Math.min(min, preSum[i - k + 1]);
            }
        }
        return max;
    }
}

python

class Solution:
    """
    @param: nums: an array of integer
    @param: k: an integer
    @return: the largest sum
    """
    def maxSubarray4(self, nums, k):
        # write your code here
        if nums is None or len(nums) == 0 or k <= 0 or len(nums) < k:
            return 0
        preSum, maxVal, minVal = [0] * (len(nums) + 1), float('-inf') ,0
        for i in range(1, len(nums) + 1):
            preSum[i] = preSum[i - 1] + nums[i - 1]
            if i >= k and maxVal < preSum[i] - minVal:
                maxVal = preSum[i] - minVal
            if i >= k:
                minVal = min(minVal, preSum[i - k + 1])
        return maxVal