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:

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

Credits:
Special thanks to @vinod23 for adding this problem and creating all test cases.

这道题的提示说了我们肯定不能把范围内的所有数字都更新，而是只更新开头结尾两个数字就行了，那么我们的做法就是在开头坐标startIndex位置加上inc，而在结束位置加1的地方加上-inc，那么根据题目中的例子，我们可以得到一个数组，nums = {-2, 2, 3, 2, -2, -3}，然后我们发现对其做累加和就是我们要求的结果result = {-2, 0, 3, 5, 3}

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

reference: http://massivealgorithms.blogspot.com/2016/06/leetcode-370-range-addition.html