[刷题]Subarray Sum Closest

最新推荐文章于 2020-08-07 17:09:48 发布

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最新推荐文章于 2020-08-07 17:09:48 发布

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分类专栏： high frequency

本文链接：https://blog.youkuaiyun.com/willshine19/article/details/48494129

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本文探讨了在数组中寻找两个子数组边界的问题，使这两个子数组的和最接近。提出了两种解决方案：一种时间复杂度为O(n^2)，另一种通过排序改进至O(nlogn)。

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[LintCode]Subarray Sum Closest

Version 1 O(n^2) 超时

public class Solution {
    /**
     * @param nums: A list of integers
     * @return: A list of integers includes the index of the first number 
     *          and the index of the last number
     */
    public ArrayList<Integer> subarraySumClosest(int[] nums) {
        // 2015-09-16 O(n^2)
        ArrayList<Integer> list = new ArrayList<>();
        if (nums == null || nums.length < 1) {
            return list;
        }
        
        int[] sum = new int[nums.length + 1];
        sum[0] = 0;
        
        int minSub = Math.abs(nums[0]);
        int start = 0;
        int end = 0;
        
        for (int i = 0; i < nums.length; i++) {
            sum[i + 1] = nums[i] + sum[i];
            for (int j = 0; j < i + 1; j++) {
                int absSub = Math.abs(sum[i + 1] - sum[j]);
                if (absSub < minSub) {
                    minSub = absSub;
                    start = j;
                    end = i;
                }
            }
        }
        
        list.add(start);
        list.add(end);
        return list;
    }
}

Version 2 O(nlogn)

public class Solution {
    /**
     * @param nums: A list of integers
     * @return: A list of integers includes the index of the first number 
     *          and the index of the last number
     */
     
    class Pair {
        int sum;
        int index;
        public Pair(int s, int i) {
            sum = s;
            index = i;
        }
    }
    
    public ArrayList<Integer> subarraySumClosest(int[] nums) {
        // write your code here
        ArrayList<Integer> res = new ArrayList<Integer> ();
        if (nums == null || nums.length == 0) {
            return res;
        } 
        
        int len = nums.length;
        if(len == 1) {
            res.add(0);
            res.add(0);
            return res;
        }
        
        Pair[] sums = new Pair[len+1];
        int prev = 0;
        sums[0] = new Pair(0, 0);
        
        for (int i = 1; i <= len; i++) {
            sums[i] = new Pair(prev + nums[i-1], i);
            prev = sums[i].sum;
        }
        // 排序
        Arrays.sort(sums, new Comparator<Pair>() {
           public int compare(Pair a, Pair b) {
               return a.sum - b.sum;
           } 
        });
        
        int ans = Integer.MAX_VALUE;
        for (int i = 1; i <= len; i++) {
            
            if (ans > sums[i].sum - sums[i-1].sum) {
                ans = sums[i].sum - sums[i-1].sum;
                res.clear();
                int[] temp = new int[] {sums[i].index - 1, sums[i - 1].index - 1};
                Arrays.sort(temp);
                res.add(temp[0] + 1);
                res.add(temp[1]);
            }
        }
        
        return res;
    }
}