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  • 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.

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  • 原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/7518802.html
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