标签：style blog color io os ar for sp div
Leetcode 加新题了
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.
整理思路不难，就是维护一个包含当前位置的最大正数和最小负数。在处理负数的时候需要注意。
1 class Solution { 2 public: 3 int maxProduct(int arr[], int n) { 4 if (n <= 0) return 0; 5 int maxPos = 1, minNeg = 1, max = INT_MIN; 6 for (int i = 0; i < n; ++i) { 7 if (arr[i] > 0) { 8 maxPos = maxPos * arr[i]; 9 minNeg = minNeg > 0 ? 1: minNeg * arr[i]; 10 if (maxPos > max) max = maxPos; 11 } else if (arr[i] == 0) { 12 maxPos = minNeg = 1; 13 if (max < 0) max = 0; 14 } else { // arr[i] < 0 15 int tmp = maxPos; 16 if (minNeg > 0) { 17 maxPos = 1; 18 } else { 19 maxPos = minNeg * arr[i]; 20 if (maxPos > max) max = maxPos; 21 } 22 minNeg = tmp * arr[i]; 23 if (minNeg > max) max = minNeg; 24 } 25 } 26 return max; 27 } 28 };
看了一下solution之后，还有更简洁的写法。我还是好弱。每天都能学到新东西啊。
1 class Solution { 2 public: 3 int maxProduct(int arr[], int n) { 4 if (n <= 0) return 0; 5 int maxSoFar = arr[0], minSoFar = arr[0], ans = arr[0]; 6 for (int i = 1; i < n; ++i) { 7 int tmp = maxSoFar; 8 maxSoFar = max(max(arr[i], maxSoFar * arr[i]), minSoFar * arr[i]); 9 minSoFar = min(min(arr[i], tmp * arr[i]), minSoFar * arr[i]); 10 if (maxSoFar > ans) ans = maxSoFar; 11 } 12 return ans; 13 } 14 };
Leetcode | Maximum Product Subarray
标签：style blog color io os ar for sp div
原文地址：http://www.cnblogs.com/linyx/p/4034938.html
