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Merge pull requestyoungyangyang04#483 from jackeyjia/patch-8

add js solution for maxProfit II

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`@@ -201,6 +201,29 @@ class Solution:`
`201`	`201`
`202`	`202`	`Go：`
`203`	`203`
	`204`	`+Javascript：`
	`205`	+```javascript
	`206`	`+constmaxProfit= (prices)=> {`
	`207`	`+let dp=Array.from(Array(prices.length), ()=>Array(2).fill(0));`
	`208`	`+// dp[i][0] 表示第i天持有股票所得现金。`
	`209`	`+// dp[i][1] 表示第i天不持有股票所得最多现金`
	`210`	`+ dp[0][0]=0- prices[0];`
	`211`	`+ dp[0][1]=0;`
	`212`	`+for(let i=1; i<prices.length; i++) {`
	`213`	`+// 如果第i天持有股票即dp[i][0]，那么可以由两个状态推出来`
	`214`	`+// 第i-1天就持有股票，那么就保持现状，所得现金就是昨天持有股票的所得现金即：dp[i - 1][0]`
	`215`	`+// 第i天买入股票，所得现金就是昨天不持有股票的所得现金减去今天的股票价格即：dp[i - 1][1] - prices[i]`
	`216`	`+ dp[i][0]=Math.max(dp[i-1][0], dp[i-1][1]- prices[i]);`
	`217`	`+`
	`218`	`+// 在来看看如果第i天不持有股票即dp[i][1]的情况，依然可以由两个状态推出来`
	`219`	`+// 第i-1天就不持有股票，那么就保持现状，所得现金就是昨天不持有股票的所得现金即：dp[i - 1][1]`
	`220`	`+// 第i天卖出股票，所得现金就是按照今天股票佳价格卖出后所得现金即：prices[i] + dp[i - 1][0]`
	`221`	`+ dp[i][1]=Math.max(dp[i-1][1], dp[i-1][0]+ prices[i]);`
	`222`	`+ }`
	`223`	`+`
	`224`	`+return dp[prices.length-1][0];`
	`225`	`+};`
	`226`	+```
`204`	`227`
`205`	`228`
`206`	`229`

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