1480. 一维数组的动态和

https://leetcode-cn.com/problems/running-sum-of-1d-array/

Java

/*
 * @Author: Goog Tech
 * @Date: 2020-08-15 12:12:07
 * @LastEditTime: 2020-08-15 12:16:24
 * @Description: https://leetcode-cn.com/problems/running-sum-of-1d-array/
 * @FilePath: \leetcode-googtech\#1480. Running Sum of 1d Array\Solution.java
 * @WebSite: https://algorithm.show/
 */

/*
 * @lc app=leetcode.cn id=1480 lang=java
 *
 * [1480] 一维数组的动态和
 */

// @lc code=start
class Solution {

    // 双指针法
    public int[] runningSum(int[] nums) {
        for(int i = 0, j = 0; j < nums.length; j++) {
            i = nums[j] + i;
            nums[j] = i;
        }
        return nums;
    }
}
// @lc code=end

Python

'''
Author: Goog Tech
Date: 2020-08-15 12:03:38
LastEditTime: 2020-08-15 12:16:49
Description: https://leetcode-cn.com/problems/running-sum-of-1d-array/
FilePath: \leetcode-googtech\#1480. Running Sum of 1d Array\Solution.py
WebSite: https://algorithm.show/
'''

#
# @lc app=leetcode.cn id=1480 lang=python
#
# [1480] 一维数组的动态和
#

# @lc code=start
class Solution(object):

    # 动态规划法
    def runningSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        # 从第二个元素开始遍历数组, 即当前位置等于当前位置的值加上上一个位置的值
        for i in range(1, len(nums)): nums[i] = nums[i - 1] + nums[i]
        return nums
# @lc code=end