这道题的最简单算法，时间复杂度O(nlogn):

    public List<Integer> targetIndices(int[] nums, int target) {
        Arrays.sort(nums);
        List<Integer> res = new ArrayList<>();
        for(int i=0;i<nums.length;i++){
            if(nums[i]==target)
                res.add(i);
        }
        return res;
    }

把查找数字那部分binary search的算法，总的时间复杂度还是O(nlogn):

    public List<Integer> targetIndices(int[] nums, int target) {
        Arrays.sort(nums);
        List<Integer> res = new ArrayList<>();
        int l = 0, r = nums.length-1;
        while(l<=r){
            int mid = (l+r)/2;
            if(nums[mid]<target){
                l = mid+1;
            }else if(nums[mid]>target){
                r = mid -1;
            }else{
                int i = mid;
                while(i>=0){
                    if(nums[i]==target){
                        res.add(0, i);
                    }
                    i--;
                }
                i = mid+1;
                while(i<nums.length){
                    if(nums[i]==target){
                        res.add(i);
                    }
                    i++;
                }
                return res;
            }
        }
        return res;
    }

不需要排序的算法，时间复杂度O(n)

public List<Integer> targetIndices(int[] nums, int target) {
         int belowTCount=0, tCount=0;    
        for(int n : nums){
            if(n<target)
                belowTCount++;
            else if(n==target)
                tCount++;
        }
        List<Integer> ans = new ArrayList<>();
            for(int t=0;t<tCount;t++)
                ans.add(belowTCount++);
    
        return ans;
    }