leetcode-26双周赛-5399-数位成本和为目标值的最大数字

Posted 2020-12-22 真不知道叫啥好

题目描述：

 提交：背包问题

class Solution:
    def largestNumber(self, cost, target: int) -> str:
        dp = [[0 for _ in range(target + 1)] for _ in range(10)]

        def tmax(a,b):
            if not a:
                return b
            elif not a:
                return b
            elif int(b) > int(a):
                return b
            else:
                return a

        for i in range(10):
            for j in range(target+1):
                if i == 0 or j == 0:
                    dp[i][j] = 0
                elif cost[i-1] > j:
                    dp[i][j] = dp[i-1][j]
                elif cost[i-1] == j:
                    dp[i][j] = tmax(dp[i-1][j],str(i))
                else:
                    tmp = 0
                    if dp[i][j-cost[i-1]]:
                        tmp = str(i)+str(dp[i][j-cost[i-1]])
                    dp[i][j] = tmax(dp[i-1][j],tmp)
        for j in range(9,-1,-1):
            if dp[j][target]:
                return dp[j][target]
        return "0"

优化：

class Solution:
    def largestNumber(self, cost: List[int], target: int) -> str:
        dp = [-1 for j in range(target + 1)]
        dp[0] = 0
        for i in range(8, -1, -1):
            for j in range(cost[i], target + 1):
                if dp[j - cost[i]] < 0:
                    continue
                dp[j] = max(dp[j], dp[j - cost[i]] * 10 + (i + 1))
        if dp[target] >= 0:
            return str(dp[target])
        else:
            return \'0\'

方法二;自底向上

class Solution:
    def largestNumber(self, cost: List[int], tar: int) -> str:
        mi = min(cost)
        @lru_cache(None)
        def dp(target): # target 下的最大值
            if target == 0: return 0
            if target < mi: return -float(\'inf\')
            res = -float(\'inf\')
            for x in range(9):
                res = max(dp(target - cost[x])*10 + x + 1, res)
            return res
        res = dp(tar)
        return str(res) if res > 0 else "0"