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Create Apply Operations to an Array

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	`1`	`+想法:用一個 index 紀錄非0的值的末尾位置,每次只要有要加入的值就加在末尾並更新位置即可`
	`2`	`+因為 index 永遠落後於 i , 所以不需擔心更新到前面的問題`
	`3`	`+`
	`4`	`+Time Complexity : O(n) for traversing the array`
	`5`	`+Space Complexity : O(n) for the answer array`
	`6`	`+`
	`7`	`+class Solution {`
	`8`	`+public:`
	`9`	`+ vector<int> applyOperations(vector<int>& nums) {`
	`10`	`+ int index = 0 ;`
	`11`	`+ for(int i = 0 ; i < nums.size() ; i++) {`
	`12`	`+ if ( nums[i] != 0 && i < nums.size() - 1 && nums[i] == nums[i + 1] ) {`
	`13`	`+ nums[index++] = nums[i] * 2 ;`
	`14`	`+ nums[i + 1] = 0 ;`
	`15`	`+ }`
	`16`	`+ else if ( nums[i] != 0 ) {`
	`17`	`+ nums[index++] = nums[i] ;`
	`18`	`+ }`
	`19`	`+ }`
	`20`	`+`
	`21`	`+ for(int i = index ; i < nums.size() ; i++)`
	`22`	`+ nums[i] = 0 ;`
	`23`	`+`
	`24`	`+ return nums ;`
	`25`	`+ }`
	`26`	`+};`

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