Maximum Subarray Difference Solution

最大子数组差值算法

最新推荐文章于 2020-05-03 13:38:30 发布

原创最新推荐文章于 2020-05-03 13:38:30 发布 · 405 阅读

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本文介绍了一种求解两个子数组间最大差值的高效算法实现。通过两次扫描数组，分别从左到右和从右到左进行，计算每个位置上的最大值和最小值，最终得出最大差值。该算法采用Java实现，并详细展示了如何计算最大子数组差值。

这道题感觉非常有意思，值得记录一下。思路来源：https://codesolutiony.wordpress.com/2015/05/28/lintcode-maximum-subarray-difference/

public class Solution {
    /**
     * @param nums: A list of integers
     * @return: An integer indicate the value of maximum difference between two
     *          Subarrays
     */
    public int maxDiffSubArrays(ArrayList<Integer> nums) {
        // write your code
        
        int len = nums.size(); 
        int[] maxLeft = new int[len], minLeft = new int[len];
        
        // sweep from left
        int max = nums.get(0), min = nums.get(0), localmax = max, localmin = max;
        maxLeft[0] = minLeft[0] = max;
        for(int i = 1; i < len; i ++){
            localmax = Math.max(nums.get(i), localmax+nums.get(i));
            localmin = Math.min(nums.get(i), localmin+nums.get(i));
            maxLeft[i] = Math.max(max, localmax);
            minLeft[i] = Math.min(min, localmin);
        }
        
        
        //sweep from right
        max = min = nums.get(len-1);
        int res = Math.max(max-minLeft[len-2], maxLeft[len-2]-min);
        localmax = localmin = nums.get(len-1);
        for(int i = len-2; i > 0; i--){
            localmax = Math.max(nums.get(i), localmax+nums.get(i));
            localmin = Math.min(nums.get(i), localmin+nums.get(i));
            int maxRight_i = Math.max(max, localmax);
            int minRight_i = Math.min(min, localmin);
            
            int tmp = Math.max(maxRight_i - minLeft[i-1], maxLeft[i-1] - minRight_i);
            res = Math.max(tmp, res);
        }
        
        return res;
    }
}