[LeetCode] 370. Range Addition 范围相加

 

Assume you have an array of length n initialized with all 0's and are given k update operations.

Each operation is represented as a triplet: [startIndex, endIndex, inc] which increments each element of subarray A[startIndex ... endIndex] (startIndex and endIndex inclusive) with inc.

Return the modified array after all k operations were executed.

Example:

Given:

    length = 5,
    updates = [
        [1,  3,  2],
        [2,  4,  3],
        [0,  2, -2]
    ]

Output:

    [-2, 0, 3, 5, 3]

Explanation:

Initial state:
[ 0, 0, 0, 0, 0 ]

After applying operation [1, 3, 2]:
[ 0, 2, 2, 2, 0 ]

After applying operation [2, 4, 3]:
[ 0, 2, 5, 5, 3 ]

After applying operation [0, 2, -2]:
[-2, 0, 3, 5, 3 ]

Hint:

  1. Thinking of using advanced data structures? You are thinking it too complicated.
  2. For each update operation, do you really need to update all elements between i and j?
  3. Update only the first and end element is sufficient.
  4. The optimal time complexity is O(k + n) and uses O(1) extra space.

Credits:
Special thanks to @vinod23 for adding this problem and creating all test cases.

 

这道题刚添加的时候我就看到了,当时只有1个提交,0个接受,于是我赶紧做,提交成功后发现我是第一个提交成功的,哈哈,头一次做沙发啊,有点小激动~ 这道题的提示说了我们肯定不能把范围内的所有数字都更新,而是只更新开头结尾两个数字就行了,那该怎么做呢?假设我们的数组范围是 [0, n),需要更新的区间是 [start, end],更新值是 inc,那么将区间 [start, end] 中每个数字加上 inc,等同于将区间 [start, n) 内的数字都加上 inc,然后将 [end+1, n) 区间内数字都减去 inc,明白了可以这样转换之后,我们还是不能每次都更新区间内所有的值,所以需要换一种标记方式,做法就是在开头坐标 start 位置加上 inc,而在结束位置 end 加1的地方加上 -inc。就比如说需要将新区间 [1, 3] 内数字都加2,那么我们在1的位置加2,在4的位置减2,于是数组就变成了 [0, 2, 0, 0, -2]。假如就只有这一个操作,如何得到最终的结果呢,答案是建立累加和数组,变成 [0, 2, 2, 2, 0],我们发现正好等同于直接将区间 [1, 3] 内的数字都加2。进一步分析,建立累加和数组的操作实际上是表示当前的数字对之后的所有位置上的数字都有影响,那么我们在 start 位置上加了2,表示在 [start, n) 区间范围内每个数字都加了2,而实际上只有 [start, end] 区间内的数字才需要加2,为了消除这种影响,我们需要将 [end+1, n) 区间内的数字都减去2,所以才在 end+1 位置上减去了2,那么建立累加和数组的时候就相当于后面所有的数字都减去了2。需要注意的是这里 end 可能等于 n-1,则 end+1 可能会越界,所以我们初始化数组的长度为 n+1,就可以避免越界了。那么根据题目中的例子,我们可以得到一个数组,nums = {-2, 2, 3, 2, -2, -3},然后对其做累加和就是我们要求的结果 result = {-2, 0, 3, 5, 3},参见代码如下:

 

解法一:

class Solution {
public:
    vector<int> getModifiedArray(int length, vector<vector<int>>& updates) {
        vector<int> res, nums(length + 1, 0);
        for (int i = 0; i < updates.size(); ++i) {
            nums[updates[i][0]] += updates[i][2];
            nums[updates[i][1] + 1] -= updates[i][2];
        }
        int sum = 0;
        for (int i = 0; i < length; ++i) {
            sum += nums[i];
            res.push_back(sum);
        }
        return res;
    }
}; 

 

我们可以在空间上稍稍优化下上面的代码,用 res 来代替 nums,最后把 res 中最后一个数字去掉即可,参见代码如下:

 

解法二:

class Solution {
public:
    vector<int> getModifiedArray(int length, vector<vector<int>>& updates) {
        vector<int> res(length + 1);
        for (auto a : updates) {
            res[a[0]] += a[2];
            res[a[1] + 1] -= a[2];
        }
        for (int i = 1; i < res.size(); ++i) {
            res[i] += res[i - 1];
        }
        res.pop_back();
        return res;
    }
}

 

Github 同步地址:

https://github.com/grandyang/leetcode/issues/370

 

类似题目:

Range Addition II

 

参考资料:

https://leetcode.com/problems/range-addition/

https://leetcode.com/problems/range-addition/discuss/84223/My-Simple-C%2B%2B-Solution

https://leetcode.com/problems/range-addition/discuss/84217/Java-O(K-%2B-N)time-complexity-Solution

https://leetcode.com/problems/range-addition/discuss/84219/Java-O(n%2Bk)-time-O(1)-space-with-algorithm-explained

 

LeetCode All in One 题目讲解汇总(持续更新中...)

posted @ 2016-06-30 08:14  Grandyang  阅读(13333)  评论(3编辑  收藏  举报
Fork me on GitHub