825. Friends Of Appropriate Ages

用presum的方法 统计有多少个小于等于这个数的年龄

 

 

class Solution {
    public int numFriendRequests(int[] ages) {
        if(ages.length <= 1) return 0;
        int[] count = new int[121];
        for(int i = 0; i < ages.length; i++){
            count[ages[i]]++;
        }
        int[] preSum = new int[122];
        for(int i = 0; i < 121; i++){
            preSum[i+1] = count[i] + preSum[i];
        }
        int res = 0;
        for(int i = 0; i < ages.length; i++){
            int max = ages[i];
            int min = ages[i] / 2 + 7;
            res += Math.max(0, preSum[max+1] - preSum[min+1] - 1); // 减去自身也要大于0啊！！！
            // System.out.println("ages:" + ages[i] + res);
        }
        return res;
    }
}