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  • 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 }
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  • 原文地址:https://www.cnblogs.com/EdwardLiu/p/6194033.html
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