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@Geekhyt
Description
栈
- 扫描过的字符入栈暂存
- 如果扫描的当前字符与栈顶元素相同,则将栈顶元素出栈
- 将栈中剩余的字符转换成字符串
const removeDuplicates = function(s) { const stack = [] for (const i of s) { if (stack.length && stack[stack.length - 1] === i) { stack.pop() } else { stack.push(i) } } return stack.join('') }
- 时间复杂度: O(n)
- 空间复杂度: O(n)