Commit 3c58ad7

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* odd-even-linked-list: add C++ implementation * move-zeros: add C++ implementation * subsets: add C++ implementation

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Lines changed: 26 additions & 1 deletion

Original file line number	Diff line number	Diff line change
`@@ -31,6 +31,9 @@ Minimize the total number of operations.`
`31`	`31`	`无`
`32`	`32`
`33`	`33`	`## 代码`
	`34`	`+* 语言支持:JS,C++`
	`35`	`+`
	`36`	`+JavaScript Code:`
`34`	`37`	```js
`35`	`38`	`/*`
`36`	`39`	`* @lc app=leetcode id=283 lang=javascript`
`@@ -79,4 +82,26 @@ var moveZeroes = function(nums) {`
`79`	`82`	`}`
`80`	`83`	`};`
`81`	`84`	```
`82`		`-`
	`85`	`+C++ Code:`
	`86`	`+> 解题思想与上面JavaScript一致,做了少许代码优化(非性能优化,因为时间复杂度都是O(n)):`
	`87`	`+> 增加一个游标来记录下一个待处理的元素的位置,这样只需要写一次循环即可。`
	`88`	+```C++
	`89`	`+class Solution {`
	`90`	`+public:`
	`91`	`+ void moveZeroes(vector<int>& nums) {`
	`92`	`+ vector<int>::size_type nonZero = 0;`
	`93`	`+ vector<int>::size_type next = 0;`
	`94`	`+ while (next < nums.size()) {`
	`95`	`+ if (nums[next] != 0) {`
	`96`	`+ // 使用std::swap()会带来8ms的性能损失`
	`97`	`+ // swap(nums[next], nums[nonZero]);`
	`98`	`+ auto tmp = nums[next];`
	`99`	`+ nums[next] = nums[nonZero];`
	`100`	`+ nums[nonZero] = tmp;`
	`101`	`+ ++nonZero;`
	`102`	`+ }`
	`103`	`+ ++next;`
	`104`	`+ }`
	`105`	`+ }`
	`106`	`+};`
	`107`	+```

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Original file line number	Diff line number	Diff line change
`@@ -49,6 +49,9 @@ Output:`
`49`	`49`
`50`	`50`	`## 代码`
`51`	`51`
	`52`	`+* 语言支持:JS,C++`
	`53`	`+`
	`54`	`+JavaScript Code:`
`52`	`55`	```js
`53`	`56`
`54`	`57`	`/*`
`@@ -104,6 +107,27 @@ var subsets = function(nums) {`
`104`	`107`	`return list;`
`105`	`108`	`};`
`106`	`109`	```
	`110`	`+C++ Code:`
	`111`	+```C++
	`112`	`+class Solution {`
	`113`	`+public:`
	`114`	`+ vector<vector<int>> subsets(vector<int>& nums) {`
	`115`	`+ auto ret = vector<vector<int>>();`
	`116`	`+ auto tmp = vector<int>();`
	`117`	`+ backtrack(ret, tmp, nums, 0);`
	`118`	`+ return ret;`
	`119`	`+ }`
	`120`	`+`
	`121`	`+ void backtrack(vector<vector<int>>& list, vector<int>& tempList, vector<int>& nums, int start) {`
	`122`	`+ list.push_back(tempList);`
	`123`	`+ for (auto i = start; i < nums.size(); ++i) {`
	`124`	`+ tempList.push_back(nums[i]);`
	`125`	`+ backtrack(list, tempList, nums, i + 1);`
	`126`	`+ tempList.pop_back();`
	`127`	`+ }`
	`128`	`+ }`
	`129`	`+};`
	`130`	+```
`107`	`131`
`108`	`132`	`## 相关题目`
`109`	`133`

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