Leetcode: 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:


Time Complexity: O(N+K), Space: O(1)

  1. For each update operation, do you really need to update all elements between i and j?
  2. Update only the first and end element is sufficient.
  3. The optimal time complexity is O(k + n) and uses O(1) extra space.
 1 public 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]+1 < length) res[update[1]+1] -= update[2];
 7         }
 8         for (int i=1; i<length; i++) {
 9             res[i] = res[i] + res[i-1];
10         }
11         return res;
12     }
13 }
原文地址:https://www.cnblogs.com/EdwardLiu/p/6194033.html