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 ]

在这里只需要对start和end操作，不需要操作start--end。这里初始化res[length]数组，res[start] + target, res[end+1] - target，最后累加res数组元素。代码如下：

public class Solution {
    public int[] getModifiedArray(int length, int[][] updates) {
        int[] res = new int[length];
        for (int[] update: updates) {
            res[update[0]] += update[2];
            if (update[1] + 1 < length)
                res[update[1] + 1] -= update[2];
        }
        for (int i = 1; i < length; i ++) {
            res[i] += res[i - 1];
        }
        return res;
    }
}