LeetCode 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 ]

Hint:

• 1. Thinking of using advanced data structures? You are thinking it too complicated.
• 1. For each update operation, do you really need to update all elements between i and j?
• 1. Update only the first and end element is sufficient.
• 1. The optimal time complexity is O(k + n) and uses O(1) extra space.

Diffculty Medium

Similar Problems

Analysis

这道题的提示说了我们肯定不能把范围内的所有数字都更新，而是只更新开头结尾两个数字就行了， 那么我们的做法就是在开头坐标startIndex位置加上inc，而在结束位置加1的地方加上-inc， 那么根据题目中的例子，我们可以得到一个数组，nums = {-2, 2, 3, 2, -2, -3}， 然后我们发现对其做累加和就是我们要求的结果result = {-2, 0, 3, 5, 3}，参见代码如下：

Solutions

解法一：

class Solution {
public:
    vector<int> getModifiedArray(int length, vector<vector<int>>& updates) {
        vector<int> res, nums(length + 1, 0);
        for (int i = 0; i < updates.size(); ++i) {
            nums[updates[i][0]] += updates[i][2];
            nums[updates[i][1] + 1] -= updates[i][2];
        }
        int sum = 0;
        for (int i = 0; i < length; ++i) {
            sum += nums[i];
            res.push_back(sum);
        }
        return res;
    }
};

解法二： 我们可以在空间上稍稍优化下上面的代码，用res来代替nums，最后把res中最后一个数字去掉即可

class Solution {
public:
    vector<int> getModifiedArray(int length, vector<vector<int>>& updates) {
        vector<int> res(length + 1);
        for (auto a : updates) {
            res[a[0]] += a[2];
            res[a[1] + 1] -= a[2];
        }
        for (int i = 1; i < res.size(); ++i) {
            res[i] += res[i - 1];
        }
        res.pop_back();
        return res;
    }
}

Reference

https://leetcode.com/discuss/111343/my-simple-c-solution