Open qappleh opened 3 years ago
解题思路 Map作哈希表,Map.values()用Set去重与Map.size相等则独一无二 代码
var uniqueOccurrences = function(arr, a) {
return a = arr.reduce((h, v) => h.set(v, (h.get(v) || 0) + 1), new Map()), new Set(a.values()).size === a.size
};
解题思路 {}或Object.create(null)作哈希表。Object.values()用Set去重与length相等则独一无二 代码 · {}
var uniqueOccurrences = function(arr, a) {
return a = Object.values(arr.reduce((h, v) => (h[v] = (h[v] || 0) + 1, h), {})), new Set(a).size === a.length
};
代码 · Object.create(null)
var uniqueOccurrences = function(arr, a) {
return a = Object.values(arr.reduce((h, v) => (h[v] = (h[v] || 0) + 1, h), Object.create(null))), new Set(a).size === a.length
};
解题思路 {}作双哈希表,前者统计出现次数,后者统计出现次数是否出现过。没出现过则独一无二
代码
var uniqueOccurrences = function(arr, h = {}, m = {}) {
arr.forEach(v => h[v] = (h[v] || 0) + 1)
for (var k in h)
if (m[h[k]]) return false
else m[h[k]] = 1
return true
};
解题思路 {}作哈希表,Object.values()用every判断lastIndexOf倒序索引与正序索引是否相等 代码
var uniqueOccurrences = function(arr) {
return Object.values(arr.reduce((h, v) => (h[v] = (h[v] || 0) + 1, h), {})).every((v, i, a) => a.lastIndexOf(v) === i)
};
解题思路 {}作哈希表,数组sort升序。前后数相同,长度+1。不同,检测长度是否已放入哈希表 sort在不同运行环境根据数组长度,有不同实现,复杂度由具体实现决定 代码
var uniqueOccurrences = function(arr, l = 1, h = {}) {
return arr.length === 0 || arr.sort((a, b) => a - b).reduce((p, c, _, a) => p === c ? (l++, c) : h[l] ? (a.splice(1), true) : (h[l] = 1, l = 1, c)) !== true && h[l] === undefined
};
如果每个数的出现次数都是独一无二的,就返回 true;否则返回 false。
示例 1:
示例 2:
示例 3:
提示: