人就是这样的，想来想去，犹豫来犹豫去，觉得自己没有准备好，勇气没攒够，其实只要迈出去了那一步，就会发现其实所有的一切，早就准备好了。——巫哲Q《撒野》

528. 按权重随机选择

• 轮盘赌

class Solution {int wsum;int[] res;public Solution(int[] w) {res = w;  wsum=0;for(int i=0;i<res.length;i++){wsum += res[i];}}public int pickIndex() {double randnum = (double)(Math.random() * wsum);int temp = res[0];for(int i=0;i<res.length;i++){if(temp<randnum){temp = temp+res[i+1];}else{return i;}}return 0;}
}

二分搜索的出现变体形式(使……最大值最小，将最大值的范围作为数组，使用左右指针进行逼近)

1011. 在 D 天内送达包裹的能力

• 想到了思路但是不敢写是什么毛病，总觉得自己是错的

class Solution {public int shipWithinDays(int[] weights, int days) {int sumnum = 0;int maxnum = 0;for(int i=0;i<weights.length;i++){sumnum += weights[i];maxnum = maxnum>weights[i]?maxnum:weights[i];}int left = maxnum;int right = sumnum;while(left<=right){if(isPackage(weights,days,(left+right)/2)){right = (left+right)/2-1;}else{left = (left+right)/2+1;}}return left;}boolean isPackage(int[] weights,int days,int lowweight){int day = 0;int sumweight=0;int i=0;while(i<weights.length){if(sumweight+weights[i]<=lowweight){sumweight += weights[i];i++;}else{day++;sumweight = 0;}}return day<days;}
}

410. 分割数组的最大值

class Solution {public int splitArray(int[] nums, int k) {int maxnum = 0;int sumnum = 0;for(int i=0;i<nums.length;i++){maxnum = maxnum>nums[i]?maxnum:nums[i];sumnum += nums[i];}int left = maxnum;int right = sumnum;while(left<=right){if(isMin((left+right)/2,nums,k)){right = (left+right)/2-1;}else{left = (left+right)/2+1;} }return left;}boolean isMin(int sum, int[] nums,int k){int s = 0;int t = 0;int i = 0;while(i<nums.length){if(s+nums[i]<=sum){s += nums[i];i++;}else{s = 0;t++;}}return t<k;}
}

875. 爱吃香蕉的珂珂

• 调试的错误是：两整数进行相除向上取整时，需要对整数先转double，然后结果再int
  int avgnum = (int)Math.ceil((double)sumnum/h);
• 结果中出现如下报错时，我修改了数据类型为long，问题解决
  

class Solution {public int minEatingSpeed(int[] piles, int h) {long sumnum=0;int maxnum=0;for(int i=0;i<piles.length;i++){maxnum = maxnum>piles[i]?maxnum:piles[i];sumnum += piles[i];}int avgnum = (int)Math.ceil((double)sumnum/h);int left = avgnum;int right = maxnum;while(left<=right){if(ismeet((left+right)/2,piles,h)){right = (left+right)/2 -1;}else{left = (left+right)/2 + 1;}}return left;}boolean ismeet(int vol,int[] piles,int h){int time = 0;for(int i=0;i<piles.length;i++){time += (int)Math.ceil((double)piles[i]/vol);}return time<=h;}
}