Lintcode 604. 滑动窗口内数的和

题目描述：
给你一个大小为n的整型数组和一个大小为k的滑动窗口，将滑动窗口从头移到尾，输出从开始到结束每一个时刻滑动窗口内的数的和。

class Solution {
public:
    /**
     * @param nums: a list of integers.
     * @param k: length of window.
     * @return: the sum of the element inside the window at each moving.
     */
    vector<int> winSum(vector<int> &nums, int k) {
        if (nums.size() < k || k <= 0) {
            return vector<int>();
        }
        
        int n = nums.size();
        vector<int> sums(n - k + 1, 0);
        for (int i = 0; i < k; ++i) {
            sums[0] += nums[i];//计算nums的前k数之和
        }
        for (int i = 1; i < n - k + 1; ++i) {
            sums[i] = sums[i - 1] - nums[i - 1] + nums[i + k - 1];
        }
        return sums;
    }
};

标签：604,窗口,nums,int,Lintcode,sums,vector,滑动,指针
来源： https://blog.csdn.net/phdongou/article/details/113833652