leedcode做题总结, 题目Maximum Subarray II & Maximum Subarray Difference

最新推荐文章于 2021-12-20 23:44:47 发布

原创最新推荐文章于 2021-12-20 23:44:47 发布 · 731 阅读

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本文介绍了一种寻找数组中最大子数组和的算法，并进一步探讨了如何求解两个不重叠子数组的最大和及绝对值最大差的问题。该算法通过预处理得到左边和右边的最大子数组和，从而有效地解决了问题。

这道题是I的拓展，有点像stock的2但是还是挺不一样的，这里是求和不是求差，所以只需要对每个点i通过I的方法分别求出左右的最大子序列然后历遍相加即可，要注意的是left[i]储存的是0-i的最大sum，而right[i]则是i+1 ~ n-1的sum，这一点需要注意。

public class Solution {
    /**
     * @param nums: A list of integers
     * @return: An integer denotes the sum of max two non-overlapping subarrays
     */
    public int maxTwoSubArrays(ArrayList<Integer> nums) {
        // write your code
        if (nums.size() == 0) {    
            return 0;    
        }  
          
        int n = nums.size();  
        int[] left=new int[n];  
        int[] right=new int[n];  
        left[0]=nums.get(0);  
        
        int local = nums.get(0);
        for (int i = 1; i < nums.size(); i++) {
            local = Math.max(nums.get(i), local + nums.get(i));
            left[i] = Math.max(local, left[i-1]);
        }
        
        right[n-2]=nums.get(n-1);  
        local = nums.get(n-1);
        for (int i = n-3; i >= 0; i--) {
            local = Math.max(nums.get(i + 1), local + nums.get(i + 1));
            right[i] = Math.max(local, right[i+1]);
        }

        int value = Integer.MIN_VALUE;
        for (int i = 0; i < n-1; i++) {    
            value = value > left[i] + right[i] ? value : left[i] + right[i];
        }  
        return value;
    }
}

第二题也不难和上面不同的是绝对值最大要求两边一个最大一个最小。

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
        if (nums.size() == 0) {    
            return 0;    
        }  
          
        int n = nums.size();  
        int[] left=new int[n];  
        int[] right=new int[n];
        int[] leftMI=new int[n];  
        int[] rightMI=new int[n];
        left[0]=nums.get(0);
        leftMI[0]=nums.get(0);
        
        int local = nums.get(0);
        int localMI = nums.get(0);
        for (int i = 1; i < nums.size(); i++) {
            local = Math.max(nums.get(i), local + nums.get(i));
            left[i] = Math.max(local, left[i-1]);
            localMI = Math.min(nums.get(i), localMI + nums.get(i));
            leftMI[i] = Math.min(localMI, leftMI[i-1]);
        }
        
        right[n-2]=nums.get(n-1); 
        rightMI[n-2]=nums.get(n-1);
        local = nums.get(n-1);
        localMI = nums.get(n-1);
        for (int i = n-3; i >= 0; i--) {
            local = Math.max(nums.get(i + 1), local + nums.get(i + 1));
            right[i] = Math.max(local, right[i+1]);
            localMI = Math.min(nums.get(i + 1), localMI + nums.get(i + 1));
            rightMI[i] = Math.min(localMI, rightMI[i+1]);
        }

        
        int value1 = Integer.MIN_VALUE;
        int value2 = Integer.MIN_VALUE;
        int value = Integer.MIN_VALUE;
        
        for (int i = 0; i < n-1; i++) {    
            value1 = value1 > Math.abs(left[i] - rightMI[i]) ? value1 : Math.abs(left[i] - rightMI[i]);
            value2 = value2 > Math.abs(leftMI[i] - right[i]) ? value2 : Math.abs(leftMI[i] - right[i]);
            value = Math.max(value, Math.max(value1, value2));
        }  
        return value;
    }
}