function removeDuplicates(nums: number[]): number {
if (nums.length === 0) {
return 0
}
let slow = 0
for (let fast = 1; fast < nums.length; fast++) {
if (nums[fast] !== nums[slow]) {
slow++
nums[slow] = nums[fast]
}
}
return slow + 1
}时间复杂度:O(n),空间复杂度:O(1)。