604 · Window Sum - LintCode

# Description

Given an array of n integers, and a moving window(size k), move the window at each iteration from the start of the array, find the sum of the element inside the window at each moving.

# Example

Example 1

Input：array = [1,2,7,8,5], k = 3
Output：[10,17,20]
Explanation：
1 + 2 + 7 = 10
2 + 7 + 8 = 17
7 + 8 + 5 = 20

# Solution

	public class Solution {
	/**
	* @param nums: a list of integers.
	* @param k: length of window.
	* @return: the sum of the element inside the window at each moving.
	*/
	public int[] winSum(int[] nums, int k) {
	if (k == 0) return new int[] {};

	int left = 0, right = k - 1, sum = 0;
	//window sum
	for (int i = 0; i < k; i++){
	sum += nums[i];
	}

	int[] res = new int[nums.length - k + 1];
	int index = 0;
	System.out.println("sum "+ sum);
	res[index++] = sum; // 讓之後的 index 為 1

	System.out.println("res[] "+ res[0]);
	while (right < nums.length - 1){
	sum -= nums[left]; //10 減去 1 [(1,) 2,7,4]
	left++; // 窗口向右移
	right++;
	sum += nums[right]; //9 加上 4
	res[index++] = sum;
	}
	return res;
	}
	}

algorithm two pointers