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  • 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:

    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 }
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  • 原文地址:https://www.cnblogs.com/amazingzoe/p/6409144.html
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