[LeetCode][easy]Rank Transform of an Array

排序加 map 可以做。

/**
Runtime: 24 ms, faster than 79.61% of Java online submissions for Rank Transform of an Array.Memory Usage: 56.1 MB, less than 100.00% of Java online submissions for Rank Transform of an Array.
*/
class Solution {public int[] arrayRankTransform(int[] arr) {if (arr == null || arr.length == 0) {return new int[0];}int[] tmp = Arrays.copyOf(arr, arr.length);Arrays.sort(tmp);HashMap<Integer, Integer> map = new HashMap<>();int index = 1;for (int i=0; i<tmp.length; i++) {if (map.get(tmp[i]) != null) {continue;}map.put(tmp[i], index++);}tmp = null;int[] result = new int[arr.length];for (int i=0; i<arr.length; i++) {result[i] = map.get(arr[i]);}return result;}
}

实际上，每个数字的次序就是数组里小于该数字的数字数量，故可以使用一个数组保存小于某个数字的数字数量，由于数组下标是有序的，所以就可以不用排序。

/**
Runtime: 4 ms, faster than 98.68% of Java online submissions for Rank Transform of an Array.Memory Usage: 49.4 MB, less than 100.00% of Java online submissions for Rank Transform of an Array.
*/
class Solution {public int[] arrayRankTransform(int[] arr) {if (arr == null || arr.length == 0) {return new int[0];}int max = Integer.MIN_VALUE;int min = Integer.MAX_VALUE;for (int i : arr) {max = i > max ? i : max;min = i< min ? i : min;}int[] priorityOfNums = new int[max - min + 1];for (int i : arr) {priorityOfNums[i - min] = 1;}for (int i=1; i<priorityOfNums.length; i++) {priorityOfNums[i] += priorityOfNums[i-1];}int[] result = new int[arr.length];for (int i=0; i<result.length; i++) {result[i] = priorityOfNums[arr[i] - min];}return result;}
}

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