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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.
思路分析
差分数组：把变量记录在差分数组上进行打标，全部打标完成后，再从第二项开始逐项求与前一项的累加值。
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public int[] getModifiedArray(int length, int[][] updates) {
if (updates == null || updates.length == 0) {
return new int[length];
}
int[] diff = new int[length];
for (int[] update : updates) {
int start = update[0];
int end = update[1];
int val = update[2];
diff[start] += val;
if (end + 1 < diff.length) {
diff[end + 1] -= val;
}
}
for (int i = 1; i < length; i++) {
diff[i] += diff[i - 1];
}
return diff;
}