1798You can construct the maximum number of continuity One question daily

topic：

1798. You can construct the maximum number of continuity

Thought：

nums = [1,4,10,3,1]1 ~ 10 All can be constructed，so11 ~ 20As well。
Traversal nums = [1,1,3,4,10]， if i<= Earlier+1

$i_0(1) <= 0+1; i_1(1) <= 1+1; i_2(3) <= 2+1; i_3(4) <= 5+1; i_4(10) <= 9+1$

, ExplainiThe previous numbers can be constructed。if i > m + 1It means that it cannot be constructed。

Code：

class Solution:
    def getMaximumConsecutive(self, coins: List[int]) -> int:
        m = 0  # At the beginning, it can only be constructed 0
        coins.sort()
        for c in coins:
            if c > m + 1:  # coins Sort，There is no comparison c Smaller
                break  # Unable to construct m+1，Continue to loop is meaningless
            m += c  # Can be constructed [0,m+c] All integer
        return m + 1  # [0,m] China m+1 An integer

func getMaximumConsecutive(coins []int) int {
	// Initialize a containing0Array
	m := 0
	sort.Ints(coins)
	for _, c := range coins {
		if c > m+1 {
			break
		}
		m += c
	}
	return m + 1
}