LeetCode Range Addition

原题链接在这里：https://leetcode.com/problems/range-addition/description/

题目：

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 ]

题解：

把每段更新都合并起来，最后走一次cumulative更新. 对于每一段跟新, 只记下开始和结尾res[start] += val, res[end+1] -= val. 最后走时从前向后cumulative赋值给当前位置就好.

Time Complexity: O(res.length + updates.length). Space: O(1).

AC Java:

 1 class Solution {
 2     public int[] getModifiedArray(int length, int[][] updates) {
 3         int [] res = new int[length];
 4         for(int [] update : updates){
 5             res[update[0]] += update[2];
 6             if(update[1] < length-1){
 7                 res[update[1]+1] -= update[2];
 8             }
 9         }
10
11         int sum = 0;
12         for(int i = 0; i<length; i++){
13             sum += res[i];
14             res[i] = sum;
15         }
16         return res;
17     }
18 }

跟上Range Addition II.