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add: Maximum Product Subarray

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`@@ -35,6 +35,7 @@ This is the solutions collection of my LeetCode submissions, most of them are pr`
`35`	`35`	`\|120\|[Triangle](https://leetcode.com/problems/triangle/) \| [JavaScript](./src/triangle/res.js)\|Medium\|`
`36`	`36`	`\|130\|[Surrounded Regions](https://leetcode.com/problems/surrounded-regions/) \| [JavaScript](./src/surrounded-regions/res.js)\|Medium\|`
`37`	`37`	`\|136\|[Single Number](https://leetcode.com/problems/single-number/) \| [JavaScript](./src/single-number/res.js)\|Easy\|`
	`38`	`+\|152\|[Maximum Product Subarray](https://leetcode.com/problems/maximum-product-subarray/) \| [JavaScript](./src/maximum-product-subarray/res.js)\|Medium\|`
`38`	`39`	`\|175\|[Combine Two Tables](https://leetcode.com/problems/combine-two-tables/)\| [SQL](./src/combine-two-tables/res.txt)\|Easy\|`
`39`	`40`	`\|176\|[Second Highest Salary](https://leetcode.com/problems/second-highest-salary/)\| [SQL](./src/second-highest-salary/res.txt)\|Easy\|`
`40`	`41`	`\|177\|[Nth Highest Salary](https://leetcode.com/problems/nth-highest-salary/)\| [SQL](./src/nth-highest-salary/res.txt)\|Medium\|`

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	`1`	`+/**`
	`2`	`+ * res.js`
	`3`	`+ * @authors Joe Jiang (hijiangtao@gmail.com)`
	`4`	`+ * @date 2017年04月13日 10:10:59`
	`5`	`+ *`
	`6`	`+ * Find the contiguous subarray within an array (containing at least one number) which has the largest product.`
	`7`	`+ * For example, given the array [2,3,-2,4],`
	`8`	`+ * the contiguous subarray [2,3] has the largest product = 6.`
	`9`	`+ *`
	`10`	`+ * Solution idea: https://discuss.leetcode.com/topic/3607/sharing-my-solution-o-1-space-o-n-running-time`
	`11`	`+ *`
	`12`	`+ * @param {number[]} nums`
	`13`	`+ * @return {number}`
	`14`	`+ */`
	`15`	`+let maxProduct = function(nums) {`
	`16`	`+ let len = nums.length;`
	`17`	`+ if (!len) {`
	`18`	`+ return 0;`
	`19`	`+ }`
	`20`	`+`
	`21`	`+ let maxPre = nums[0],`
	`22`	`+ maxNow = nums[0],`
	`23`	`+ minPre = nums[0];`
	`24`	`+ for (let i = 1; i < len; i++) {`
	`25`	`+ let maxVal = Math.max(maxPre * nums[i], minPre * nums[i], nums[i]),`
	`26`	`+ minVal = Math.min(minPre * nums[i], maxPre * nums[i], nums[i]);`
	`27`	`+`
	`28`	`+ maxNow = Math.max(maxVal, maxNow);`
	`29`	`+ maxPre = maxVal;`
	`30`	`+ minPre = minVal;`
	`31`	`+ }`
	`32`	`+`
	`33`	`+ return maxNow;`
	`34`	`+};`

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