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:

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

Explanation:

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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.
    The optimal time complexity is O(k + n) and uses O(1) extra space.

譬如数组nums[n], 修改(i, j, a). 只修改nums[i] = a, nums[j + 1] = -a，就可以了。前缀和数组就是所求结果。 

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

Ref: https://discuss.leetcode.com/topic/49691/java-o-k-n-time-complexity-solution