leetcode-2089-easy

Find Target Indices After Sorting Array

You are given a 0-indexed integer array nums and a target element target.

A target index is an index i such that nums[i] == target.

Return a list of the target indices of nums after sorting nums in non-decreasing order. If there are no target indices, return an empty list. The returned list must be sorted in increasing order.

Example 1:

Input: nums = [1,2,5,2,3], target = 2
Output: [1,2]
Explanation: After sorting, nums is [1,2,2,3,5].
The indices where nums[i] == 2 are 1 and 2.
Example 2:

Input: nums = [1,2,5,2,3], target = 3
Output: [3]
Explanation: After sorting, nums is [1,2,2,3,5].
The index where nums[i] == 3 is 3.
Example 3:

Input: nums = [1,2,5,2,3], target = 5
Output: [4]
Explanation: After sorting, nums is [1,2,2,3,5].
The index where nums[i] == 5 is 4.

思路一：对数字进行完全分类，只可能出现下面三种情况，只要统计等于和小于 target 的个数就能求出解

• 大于 target
• 等于 target
• 小于 target

public List<Integer> targetIndices(int[] nums, int target) {
    int minCount = 0;
    int equalCount = 0;
    for (int i = 0; i < nums.length; i++) {
        if (nums[i] < target) {
            minCount++;
        } else if (nums[i] == target) {
            equalCount++;
        }
    }

    List<Integer> list = new ArrayList<>();
    for (int i = 0; i < equalCount; i++) {
        list.add(minCount++);
    }

    return list;
}