875. Koko Eating Bananas

Link

Solution

卡在如何決定feasible，答案很簡單, 就計算每個pile要花多少時間，然後加總之後看有沒有超過H

    public int minEatingSpeed(int[] piles, int H) {
        int left = 1;
        int right = 0;
        for(int pile : piles){
            right = Math.max(right, pile);
        }
        
        while( right > left ){
            int mid = left + (right-left)/2;
            if(isfeasible(mid, piles, H)){
                right = mid;
            }
            else{
                left = mid+1;
            }
        }
        return left;
    }
    
    public boolean isfeasible(int consume, int[] piles, int H){
        int day = 0;
        
        for(int pile : piles){
            day += pile / consume + ((pile%consume == 0)? 0 : 1);
        }
        return day <= H;
    }