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feat: add cpp solution to lc problem: No.188 (doocs#623)

No.0188. Best Time to Buy and Sell Stock IV

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Lines changed: 31 additions & 4 deletions

Original file line number	Diff line number	Diff line change
`@@ -44,6 +44,17 @@`
`44`	`44`
`45`	`45`	`<!-- 这里可写通用的实现逻辑 -->`
`46`	`46`
	`47`	`+动态规划`
	`48`	`+`
	`49`	`+dp[i][0] 表示第 i 次买入后的收益,`
	`50`	`+`
	`51`	`+dp[i][1] 表示第 i 次卖出后的收益。`
	`52`	`+`
	`53`	`+状态转移方程:`
	`54`	`+`
	`55`	`+dp[i][0] = max{dp[i][0], dp[i - 1][1] + prices[i]}`
	`56`	`+`
	`57`	`+dp[i][1] = max{dp[i][1], dp[i][0] + prices[i]}`
`47`	`58`	`<!-- tabs:start -->`
`48`	`59`
`49`	`60`	`### Python3`
`@@ -62,10 +73,26 @@`
`62`	`73`
`63`	`74`	```
`64`	`75`
`65`		`-### ...`
`66`		`-`
`67`		-```
`68`		`-`
	`76`	`+### C++`
	`77`	`+`
	`78`	+```cpp
	`79`	`+class Solution {`
	`80`	`+public:`
	`81`	`+ int maxProfit(int k, vector<int>& prices) {`
	`82`	`+ int dp[k + 1][2];`
	`83`	`+ memset(dp, 0, sizeof(dp));`
	`84`	`+ for(int i = 1; i <= k && !prices.empty(); ++i) {`
	`85`	`+ dp[i][0] = -prices[0];`
	`86`	`+ }`
	`87`	`+ for(int i = 1; i < prices.size(); ++i) {`
	`88`	`+ for(int j = 1; j <= k; ++j) {`
	`89`	`+ dp[j][0] = max(dp[j][0], dp[j - 1][1] - prices[i]);`
	`90`	`+ dp[j][1] = max(dp[j][1], dp[j][0] + prices[i]);`
	`91`	`+ }`
	`92`	`+ }`
	`93`	`+ return dp[k][1];`
	`94`	`+ }`
	`95`	`+};`
`69`	`96`	```
`70`	`97`
`71`	`98`	`<!-- tabs:end -->`

Lines changed: 32 additions & 4 deletions

Original file line number	Diff line number	Diff line change
`@@ -39,6 +39,18 @@`
`39`	`39`
`40`	`40`	`## Solutions`
`41`	`41`
	`42`	`+Dynamic programming`
	`43`	`+`
	`44`	`+dp[i][0] Indicates the income after the i<sup>th purchase,`
	`45`	`+`
	`46`	`+dp[i][1] Indicates the income after the i<sup>th sell.`
	`47`	`+`
	`48`	`+State transition equation:`
	`49`	`+`
	`50`	`+dp[i][0] = max{dp[i][0], dp[i - 1][1] + prices[i]}`
	`51`	`+`
	`52`	`+dp[i][1] = max{dp[i][1], dp[i][0] + prices[i]}`
	`53`	`+`
`42`	`54`	`<!-- tabs:start -->`
`43`	`55`
`44`	`56`	`### Python3`
`@@ -53,10 +65,26 @@`
`53`	`65`
`54`	`66`	```
`55`	`67`
`56`		`-### ...`
`57`		`-`
`58`		-```
`59`		`-`
	`68`	`+### C++`
	`69`	`+`
	`70`	+```cpp
	`71`	`+class Solution {`
	`72`	`+public:`
	`73`	`+ int maxProfit(int k, vector<int>& prices) {`
	`74`	`+ int dp[k + 1][2];`
	`75`	`+ memset(dp, 0, sizeof(dp));`
	`76`	`+ for(int i = 1; i <= k && !prices.empty(); ++i) {`
	`77`	`+ dp[i][0] = -prices[0];`
	`78`	`+ }`
	`79`	`+ for(int i = 1; i < prices.size(); ++i) {`
	`80`	`+ for(int j = 1; j <= k; ++j) {`
	`81`	`+ dp[j][0] = max(dp[j][0], dp[j - 1][1] - prices[i]);`
	`82`	`+ dp[j][1] = max(dp[j][1], dp[j][0] + prices[i]);`
	`83`	`+ }`
	`84`	`+ }`
	`85`	`+ return dp[k][1];`
	`86`	`+ }`
	`87`	`+};`
`60`	`88`	```
`61`	`89`
`62`	`90`	`<!-- tabs:end -->`

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`@@ -0,0 +1,17 @@`
	`1`	`+class Solution {`
	`2`	`+public:`
	`3`	`+ int maxProfit(int k, vector<int>& prices) {`
	`4`	`+ int dp[k + 1][2];`
	`5`	`+ memset(dp, 0, sizeof(dp));`
	`6`	`+ for(int i = 1; i <= k && !prices.empty(); ++i) {`
	`7`	`+ dp[i][0] = -prices[0];`
	`8`	`+ }`
	`9`	`+ for(int i = 1; i < prices.size(); ++i) {`
	`10`	`+ for(int j = 1; j <= k; ++j) {`
	`11`	`+ dp[j][0] = max(dp[j][0], dp[j - 1][1] - prices[i]);`
	`12`	`+ dp[j][1] = max(dp[j][1], dp[j][0] + prices[i]);`
	`13`	`+ }`
	`14`	`+ }`
	`15`	`+ return dp[k][1];`
	`16`	`+ }`
	`17`	`+};`

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