LeetCode 370. Range Addition

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:

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

思路：
1. 最直接粗暴的方法是：对k次操作，每一次都遍历n长的vector,这样复杂度o(nk)。但是，考虑到每个操作，比如[1,3,2]是对1到3这个区间的数加2，增加的都是相同的数，因此，没必要一个一个的去增加这个区间的数，一定存在更快的方法！
2. 需要提取记录的信息：把k次操作对应的首尾坐标，标记在vector上！最后复杂度o(k+n)
2. 先对给定的K次操作预处理：把每次操作的开始和结束坐标标记在n长的vector上， 比如：[1,3,2],就在index=1处加2，index=4减2，把所有的k次操作都这么处理，然后遍历一遍vecotor，index=i的位置上的最终值等于vector[0,i]这个区间上所有值累加！！
3. 例如： length = 5,
updates = [
[1, 3, 2],
[2, 4, 3],
[0, 2, -2]
]
initially, nums=[0,0,0,0,0];
after [1,3,2], nums=[0,2,0,0,-2];
after [2,4,3], nums=[0,2,3,0,-2];
after [0,2,-2],nums=[-2,2,3,2,-2]。所以，最终累加结果[-2,0,3,5,3].

vector<int> getModifiedArray(int length, vector<vector<int>>& updates) {    vector<int> res(length,0);    for(auto&op:updates){//预处理        res[op[0]]+=op[2];        if(op[1]+1<length)            res[op[1]+1]-=op[2];    }    for(int i=1;i<length;i++){//累加        res[i]+=res[i-1];    }    return res;}