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 ]

思路：是根据hint得来的，我觉得面试的时候，很难想到这种方法；

具体解释就是：不用update range里面的所有元素，把所有赋值update的过程，全部用最后累加的时候来计算，往后延伸，

startindex的时候赋值inc，在endindex+1的时候赋值-inc，目的是为了值往后累加的时候，到endindex+1的时候就不往后传递了。终止符。

比如说加个1，000100000-10000， 然后往后累加，1就会往后传递，直到-1的时候就为0，不往后传递了。

public class Solution {
    public int[] getModifiedArray(int length, int[][] updates) {
        int[] array = new int[length];
        for(int i=0; i<updates.length; i++){
            int start = updates[i][0];
            int end = updates[i][1];
            int k = updates[i][2];
            array[start] +=k;
            if(end+1<array.length){
                array[end+1] -= k;
            }
        }
        for(int i=1; i<array.length; i++){
            array[i] += array[i-1];
        }
        return array;
    }
}