【leetcode】1413. Minimum Value to Get Positive Step by Step Sum

标签：init   cal   def   elements   xpl   bsp   constrain   tput   for   

题目如下：

Given an array of integers nums, you start with an initial positive value startValue.

In each iteration, you calculate the step by step sum of startValue plus elements in nums (from left to right).

Return the minimum positive value of startValue such that the step by step sum is never less than 1. 

Example 1:

Input: nums = [-3,2,-3,4,2]
Output: 5
Explanation: If you choose startValue = 4, in the third iteration your step by step sum is less than 1.
                step by step sum
                startValue = 4 | startValue = 5 | nums
                  (4 -3 ) = 1  | (5 -3 ) = 2    |  -3
                  (1 +2 ) = 3  | (2 +2 ) = 4    |   2
                  (3 -3 ) = 0  | (4 -3 ) = 1    |  -3
                  (0 +4 ) = 4  | (1 +4 ) = 5    |   4
                  (4 +2 ) = 6  | (5 +2 ) = 7    |   2

Example 2:

Input: nums = [1,2]
Output: 1
Explanation: Minimum start value should be positive. 

Example 3:

Input: nums = [1,-2,-3]
Output: 5

Constraints:

• 1 <= nums.length <= 100
• -100 <= nums[i] <= 100

解题思路：从左往右依次累加nums的和，如果遇到和为负数的情况，只要保证 startValue = min(负数和） + 1 即可；如果全为正数，startValue = 1。

代码如下：

class Solution(object):
    def minStartValue(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        min_val = float(‘inf‘)
        amount = 0
        for i in nums:
            amount += i
            min_val = min(min_val,amount)
        if min_val >= 1:return 1
        else:return abs(min_val) + 1

【leetcode】1413. Minimum Value to Get Positive Step by Step Sum

标签：init   cal   def   elements   xpl   bsp   constrain   tput   for   

原文地址：https://www.cnblogs.com/seyjs/p/12985246.html