1）所有人分1个candy
2）从左往右查看，若满足左规则，令 candy[i] =candy[i - 1] + 1
3）从右往左查看，若满足右规则，令 candy[j] =Math.max(candy[j + 1] + 1,candy[j])，取最大值是为了在满足右规则的时候不破坏左规则

class Solution {public int candy(int[] ratings) {if(ratings.length==0) return 0;int candy[] =new int[ratings.length];int total=0;candy[0]=1;//从左向右遍历数组使其满足左规则for(int i=1;i<ratings.length;i++){candy[i]=1;if(ratings[i]>ratings[i-1]){candy[i]=candy[i-1]+1;}}//从右向左遍历数组使其满足左规则for(int i=ratings.length-2;i>=0;i--){if(ratings[i]>ratings[i+1]){candy[i]=Math.max(candy[i+1]+1,candy[i]);}}//遍历获得糖果总数for(int i=0;i<ratings.length;i++){total+=candy[i];}return total;}
}