Range Addition

标签：mat   complex   int   code   etc   struct   advance   ++   nat   

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.

 1 public class Solution {
 2     public int[] getModifiedArray(int length, int[][] updates) {
 3         int[] temp = new int[length];
 4         for (int[] update : updates) {
 5             int start = update[0], end = update[1] + 1, inc = update[2];
 6             
 7             temp[start] += inc;
 8             if (end < length)
 9                 temp[end] -= inc;
10         }
11         
12         int[] result = new int[length];
13         int sum = 0;
14         for (int i = 0; i < length; i++) {
15             result[i] = sum + temp[i];
16             sum = result[i];
17         }
18         
19         return result;
20     }
21 }

Range Addition

标签：mat   complex   int   code   etc   struct   advance   ++   nat   

原文地址：http://www.cnblogs.com/amazingzoe/p/6409144.html