Day10 daily 3 questions (1): sum gradually to get the minimum value of positive numbers

subject ：

Give you an array of integers nums . You can choose any   Positive numbers startValue As an initial value .

You need to traverse from left to right nums  Array , And will startValue Add up in sequence  nums  The values in the array .

Please make sure that the cumulative sum is always greater than or equal to 1 Under the premise of , Choose the smallest   Positive numbers   As startValue .

Example 1：

Input ：nums = [-3,2,-3,4,2]
Output ：5
explain ： If you choose startValue = 4, At the third cumulation , And less than 1 .
Sum by accumulation
  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
explain ： The smallest startValue Positive number expected .
Example 3：

Input ：nums = [1,-2,-3]
Output ：5

Tips ：

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

Ideas ：

because startValue Positive number , therefore startValue>0, And because every element of the accumulation must >0, therefore startValue The initial value of can be
Math.max(1,-nums[0]+1);
Traverse nums, When you add up and <=0 when ,start+=1-sum, At this time, the cumulative sum is also more 1-sum, namely sum+=1-sum, namely sum=1;

java Code ：

class Solution {
     public int minStartValue(int[] nums) {
        int start = Math.max(1, -nums[0] + 1);
        int sum = start;
        for (int i : nums) {
            sum +=  i;
            if (sum <= 0) {
                start += 1 - sum;
                sum = 1;
            }
        }
        return start;
    }
}