[LeetCode]Count of Smaller Numbers After Self

二分搜索，边界条件要注意

public class Solution {
    public List<Integer> countSmaller(int[] nums) {
        List<Integer> sorted = new ArrayList<Integer>();
        int length = nums.length;
        List<Integer> result = new LinkedList<Integer>();
        for (int i = length - 1; i >= 0; i--) {
            int num = helper(sorted, nums[i]);
            result.add(0, num);
        }
        return result;
    }
    private int helper(List<Integer> sorted, int num) {
        int left = 0;
        int size = sorted.size();
        if (size == 0) {
            sorted.add(num);
            return 0;
        } else if (num > sorted.get(size - 1)) {
            sorted.add(size, num);
            return size;
        }
        int right = size - 1;
        while (left + 1 < right) {
            int mid = left + (right - left) / 2;
            if (sorted.get(mid) >= num) {
                right = mid;
            } else {
                left = mid;
            }
        }
        int tmp = sorted.get(left) >= num ? left : right;
        sorted.add(tmp, num);
        return tmp;
    }
}