Docker Compose 部署 MySQL 多实例 日常运维全指南-补充
2025/12/28 14:45:55
算法原理:
解法:贪心
1ms击败98.58%
时间复杂度O(1)
无非就三种情况
①全用cost1和cost2
②挑need1或者need2最小的用costBoth覆盖掉,剩下那个单独算
③全用costBoth覆盖
三者取个最小的即可
class Solution { public long minimumCost(int c1, int c2, int c, int n1, int n2) { if(n1==0&&n2==0) return 0; long kmax=Math.max(n1,n2); long kmin=Math.min(n1,n2); long ca=(long)n1*c1+(long)n2*c2; long cb=(long)kmin*c+(long)(n1-kmin)*c1+(long)(n2-kmin)*c2; long cc=(long)kmax*c; return Math.min(ca,Math.min(cb,cc)); } }