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 }