LeetCode 152. Maximum Product Subarray

152. Maximum Product Subarray
Find the contiguous subarray within an array (containing at least one number) which has the largest product.

For example, given the array [2,3,-2,4],
the contiguous subarray [2,3] has the largest product = 6.

题目大意：给一个整型数组，求该数组中所有连续子数组的元素乘积的最大值～
分析：需要保存临时最大值和最小值，因为最大值乘以一个正数可能构成新的最大值，而最小值乘以负数也可能构成新的最大值。
result是要求的结果，maxValue为nums[i]之前的，和nums[i]相邻的乘积的最大值，minValue为nums[i]之前的，和nums[i]相邻的乘积的最小值。
首先令result、maxValue和minValue都为nums[0], i从nums[1]开始一直到结束，tempMax为考虑是否选择之前的maxValue与nums[i]相乘，如果相乘结果更大就保留，否则就选择nums[i]本身为最大。tempMin同理～
然后maxValue和minValue比较tempMax/tempMin与minValue * nums[i]的大小关系，maxValue取较大值，minValue取较小值～
而result是取所有maxValue中的最大值～最后返回result～

class Solution {
public:int maxProduct(vector<int>& nums) {if (nums.size() == 0) return 0;int result = nums[0], maxValue = nums[0], minValue = nums[0];for (int i = 1; i < nums.size(); i++) {int tempMax = max(nums[i], maxValue * nums[i]);int tempMin = min(nums[i], maxValue * nums[i]);maxValue = max(tempMax, minValue * nums[i]);minValue = min(tempMin, minValue * nums[i]);result = max(maxValue, result);}return result;}
};

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