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package leetcode_1To300;
/**
* 本代码来自 Cspiration,由 @Cspiration 提供
* 题目来源:http://leetcode.com
* - Cspiration 致力于在 CS 领域内帮助中国人找到工作,让更多海外国人受益
* - 现有课程:Leetcode Java 版本视频讲解(1-900题)(上)(中)(下)三部
* - 算法基础知识(上)(下)两部;题型技巧讲解(上)(下)两部
* - 节省刷题时间,效率提高2-3倍,初学者轻松一天10题,入门者轻松一天20题
* - 讲师:Edward Shi
* - 官方网站:https://cspiration.com
* - 版权所有,转发请注明出处
*/
public class _370_RangeAddition {
/**
* 370. 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 ]
time : O(K + N);
space : O(n)
* @param length
* @param updates
* @return
*/
public int[] getModifiedArray(int length, int[][] updates) {
int[] res = new int[length];
for (int[] update : updates) {
int value = update[2];
int start = update[0];
int end = update[1];
res[start] += value;
if (end + 1 < length) {
res[end + 1] -= value;
}
}
for (int i = 1; i < length; i++) {
res[i] += res[i - 1];
}
return res;
}
}