53. Maximum Subarray

最新推荐文章于 2025-05-15 16:32:28 发布

转载最新推荐文章于 2025-05-15 16:32:28 发布 · 54 阅读

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原文链接：http://www.cnblogs.com/jessie2009/p/9799356.html

本文深入探讨了寻找最大子数组和的算法实现，包括动态规划（DP）和分治法两种方法。通过实例[-2,1,-3,4,-1,2,1,-5,4]展示了算法的具体应用过程，最终得出最大子数组和为6。文章提供了两种算法的时间复杂度和空间复杂度分析，其中动态规划的时间复杂度为O(n)，空间复杂度为O(1)，而分治法的时间复杂度为O(n!)，空间复杂度同样为O(1)。

Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.

Example:

Input: [-2,1,-3,4,-1,2,1,-5,4],
Output: 6
Explanation: [4,-1,2,1] has the largest sum = 6.

Follow up:

If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.

//DP: Time: O(n), Space:O(1)
    public int maxSubArray(int[] nums) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        
        int sum = nums[0];
        int max = nums[0];
        
        for (int i = 1; i < nums.length; i++) {
            sum = Math.max(sum + nums[i], nums[i]);
            max = Math.max(max, sum);
        }
        
        return max;
    }

//Divide and Concur: Time: O(n!), Space: O(1)
    public int maxSubArray(int[] nums) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        
        return helper(nums, 0, nums.length - 1);
    }
    
    private int helper(int[] nums, int start, int end) {
        if (start >= end) {
            return nums[start];
        }
        
        int mid = start + (end - start) / 2;
        int mmax = nums[mid];
        int lmax = helper(nums, start, mid - 1);
        int rmax = helper(nums, mid + 1, end);
        
        int sum = mmax;
        
        for (int i = mid - 1; i >= start; --i) {
            sum += nums[i];
            mmax = Math.max(mmax, sum);
        }
        
        sum = mmax;//不要忘记计算完左面的最大值，给sum归位
        
        for (int i = mid + 1; i <= end; i++) {
            sum += nums[i];
            mmax = Math.max(sum, mmax);
        }
        
        return Math.max(mmax, Math.max(lmax, rmax));
    }

转载于:https://www.cnblogs.com/jessie2009/p/9799356.html