文、意如
題目精簡說明:
有一個排序好的陣列，需移除重複的值，只能在現有的nums陣列中存入結果，回傳陣列的長度
範例： [1,1,2] 去除重複的1之後，剩下[1,2]，回傳陣列的長度2。
原始題目:
Given an integer array nums sorted in non-decreasing order, remove the duplicates in-place such that each unique element appears only once. The relative order of the elements should be kept the same.
Since it is impossible to change the length of the array in some languages, you must instead have the result be placed in the first part of the array nums. More formally, if there are k elements after removing the duplicates, then the first k elements of nums should hold the final result. It does not matter what you leave beyond the first k elements.
Return k after placing the final result in the first k slots of nums.
Do not allocate extra space for another array. You must do this by modifying the input array in-place with O(1) extra memory.
範例參考:
Example 1:
Input: nums = [1,1,2] Output: 2, nums = [1,2,_] Explanation: Your function should return k = 2, with the first two elements of nums being 1 and 2 respectively. It does not matter what you leave beyond the returned k (hence they are underscores).
Example 2:
Input: nums = [0,0,1,1,1,2,2,3,3,4] Output: 5, nums = [0,1,2,3,4,_,_,_,_,_] Explanation: Your function should return k = 5, with the first five elements of nums being 0, 1, 2, 3, and 4 respectively. It does not matter what you leave beyond the returned k (hence they are underscores).
解題說明:
1.先預設tmp元素為0個
2.把結果先暫存到 nums[tmp] 中
3.如果沒重複時將 tmp 中的元素 +1
4.並把元素存入 nums[tmp]
5.元素最少會有一個(第一個存入的元素) ，所以最後tmp要記得在加1
完整程式碼:
/**
* @param {number[]} nums
* @return {number}
*/
var removeDuplicates = function(nums) {
tmp=0; //預設元素為0個
for(i=0;i<nums.length;i++){
if(nums[tmp]!==nums[i]){ //如果num[tmp]沒有重複元素時
tmp+=1; //將元素+1
nums[tmp]=nums[i] //並把元素存入num[tmp]
}
}
tmp+=1; //元素最少會有一個(第一個存入的元素)
return tmp
};
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