|
| 1 | +# 451. 根据字符出现频率排序 |
| 2 | + |
| 3 | +https://leetcode-cn.com/problems/sort-characters-by-frequency/ |
| 4 | + |
| 5 | +- [451. 根据字符出现频率排序](#451-根据字符出现频率排序) |
| 6 | + - [题目描述](#题目描述) |
| 7 | + - [方法1:哈希表+堆](#方法1哈希表堆) |
| 8 | + - [思路](#思路) |
| 9 | + - [复杂度分析](#复杂度分析) |
| 10 | + - [代码](#代码) |
| 11 | + |
| 12 | +## 题目描述 |
| 13 | + |
| 14 | +``` |
| 15 | +给定一个字符串,请将字符串里的字符按照出现的频率降序排列。 |
| 16 | +
|
| 17 | +示例 1: |
| 18 | +
|
| 19 | +输入: |
| 20 | +"tree" |
| 21 | +
|
| 22 | +输出: |
| 23 | +"eert" |
| 24 | +
|
| 25 | +解释: |
| 26 | +'e'出现两次,'r'和't'都只出现一次。 |
| 27 | +因此'e'必须出现在'r'和't'之前。此外,"eetr"也是一个有效的答案。 |
| 28 | +示例 2: |
| 29 | +
|
| 30 | +输入: |
| 31 | +"cccaaa" |
| 32 | +
|
| 33 | +输出: |
| 34 | +"cccaaa" |
| 35 | +
|
| 36 | +解释: |
| 37 | +'c'和'a'都出现三次。此外,"aaaccc"也是有效的答案。 |
| 38 | +注意"cacaca"是不正确的,因为相同的字母必须放在一起。 |
| 39 | +示例 3: |
| 40 | +
|
| 41 | +输入: |
| 42 | +"Aabb" |
| 43 | +
|
| 44 | +输出: |
| 45 | +"bbAa" |
| 46 | +
|
| 47 | +解释: |
| 48 | +此外,"bbaA"也是一个有效的答案,但"Aabb"是不正确的。 |
| 49 | +注意'A'和'a'被认为是两种不同的字符。 |
| 50 | +
|
| 51 | +来源:力扣(LeetCode) |
| 52 | +链接:https://leetcode-cn.com/problems/sort-characters-by-frequency |
| 53 | +著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。 |
| 54 | +``` |
| 55 | + |
| 56 | +## 方法1:哈希表+堆 |
| 57 | + |
| 58 | +### 思路 |
| 59 | + |
| 60 | +- 用哈希表来记录每个字符的出现次数 |
| 61 | +- 以字符出现次数建立一个大顶堆 |
| 62 | +- 一边弹出堆顶,一边构建新的字符串 |
| 63 | + |
| 64 | +### 复杂度分析 |
| 65 | + |
| 66 | +- 时间复杂度:$O(n+klogk)$,n 是字符串的长度,k 是字符串中字符集的大小。 |
| 67 | +- 空间复杂度:$O(k)$,k 是字符串中字符集的大小,堆的大小。 |
| 68 | + |
| 69 | +### 代码 |
| 70 | + |
| 71 | +JavaScript Code |
| 72 | + |
| 73 | +```js |
| 74 | +/** |
| 75 | + * @param {string} s |
| 76 | + * @return {string} |
| 77 | + */ |
| 78 | +var frequencySort = function (s) { |
| 79 | + const map = {} |
| 80 | + |
| 81 | + for (let i = 0; i < s.length; i++) { |
| 82 | + const c = s[i]; |
| 83 | + map[c] = (map[c] || 0) + 1 |
| 84 | + } |
| 85 | + |
| 86 | + // 堆的数据结构 [char, count] |
| 87 | + const list = Object.keys(map).map(c => [c, map[c]]) |
| 88 | + const heap = new MaxHeap(list, function comparator(inserted, compared) { |
| 89 | + return inserted[1] < compared[1]; |
| 90 | + }); |
| 91 | + |
| 92 | + let str = '' |
| 93 | + while (heap.size() > 0) { |
| 94 | + const [char, cnt] = heap.pop(); |
| 95 | + str += char.repeat(cnt) |
| 96 | + } |
| 97 | + return str |
| 98 | +}; |
| 99 | + |
| 100 | +// ************************************************** |
| 101 | + |
| 102 | +class Heap { |
| 103 | + constructor(list = [], comparator) { |
| 104 | + this.list = list; |
| 105 | + this.comparator = comparator; |
| 106 | + |
| 107 | + this.init(); |
| 108 | + } |
| 109 | + |
| 110 | + init() { |
| 111 | + const size = this.size(); |
| 112 | + for (let i = Math.floor(size / 2) - 1; i >= 0; i--) { |
| 113 | + this.heapify(this.list, size, i); |
| 114 | + } |
| 115 | + } |
| 116 | + |
| 117 | + insert(n) { |
| 118 | + this.list.push(n); |
| 119 | + const size = this.size(); |
| 120 | + for (let i = Math.floor(size / 2) - 1; i >= 0; i--) { |
| 121 | + this.heapify(this.list, size, i); |
| 122 | + } |
| 123 | + } |
| 124 | + |
| 125 | + peek() { |
| 126 | + return this.list[0]; |
| 127 | + } |
| 128 | + |
| 129 | + pop() { |
| 130 | + const last = this.list.pop(); |
| 131 | + if (this.size() === 0) return last; |
| 132 | + const returnItem = this.list[0]; |
| 133 | + this.list[0] = last; |
| 134 | + this.heapify(this.list, this.size(), 0); |
| 135 | + return returnItem; |
| 136 | + } |
| 137 | + |
| 138 | + size() { |
| 139 | + return this.list.length; |
| 140 | + } |
| 141 | +} |
| 142 | + |
| 143 | +class MaxHeap extends Heap { |
| 144 | + constructor(list, comparator) { |
| 145 | + super(list, comparator); |
| 146 | + } |
| 147 | + |
| 148 | + heapify(arr, size, i) { |
| 149 | + let largest = i; |
| 150 | + const left = Math.floor(i * 2 + 1); |
| 151 | + const right = Math.floor(i * 2 + 2); |
| 152 | + |
| 153 | + if (left < size && this.comparator(arr[largest], arr[left])) |
| 154 | + largest = left; |
| 155 | + if (right < size && this.comparator(arr[largest], arr[right])) |
| 156 | + largest = right; |
| 157 | + |
| 158 | + if (largest !== i) { |
| 159 | + [arr[largest], arr[i]] = [arr[i], arr[largest]]; |
| 160 | + this.heapify(arr, size, largest); |
| 161 | + } |
| 162 | + } |
| 163 | +} |
| 164 | +``` |
| 165 | + |
| 166 | +更多题解可以访问:[https://github.com/suukii/91-days-algorithm](https://github.com/suukii/91-days-algorithm) |
0 commit comments