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 ]

Solution：在start置value，再积分处理

思路：
对于 [startIndex, endIndex, inc_value]
在startIndex上存inc_value，endIndex+1上存-1 * inc_value
返回结果时遍历累计sum(积分)

for example it will look like:
[1 , 3 , 2] , [2, 3, 3] (length = 5)
res[ 0, 2, ,0, 0 -2 ]
res[ 0 ,2, 3, 0, -5]
sum 0, 2, 5, 5, 0
res[0, 2, 5, 5, 0]

Time Complexity: O(k+N) Space Complexity: O(1)

Solution Code：

class Solution {
    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;
    }
}