Description:
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:
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 ]
Link:
https://leetcode.com/problems/range-addition/description/
解题方法：
除了直接按照题意暴力破解，还有一种使用加法结合律的方法，以以上例子为例：
当有只有一个命令：[1, 3, 2]
- 操作1
在start位置进行+= 2的操作，在end+1的位置进行-=2的操作。 - 操作2
当我们遍历整个数组进行a[i] += a[i - 1]的操作时，即可得到我们想要的数组。
这道题所有的操作都是加法，所以不管有多少个命令，我们都先进行操作1，然后再最后统一进行操作2，即可得到所求的数组。
Tips:
O(length + updates.size())
Time Complexity：
完整代码：
vector<int> getModifiedArray(int length, vector<vector<int>>& updates)
{
vector<int> result(length);
for(auto a: updates)
{
result[a[0]] += a[2];
if(a[1] + 1 < length)
result[a[1] + 1] -= a[2];
}
for(int i = 1; i < length; i++)
result[i] += result[i - 1];
return result;
}
