leetcode:Maximum Product Subarray
2014-09-26 12:43
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简单的DP题目，不过要注意负负得正这个就是了，<span style="font-size:18px;">public class Solution {
public int maxProduct(int[] A){
int ans = 0;
if(A != null && A.length != 0){
int maxArray[] = new int[A.length];
int minArray[] = new int[A.length];
ans = minArray[0] = maxArray[0] = A[0];
for(int i = 1; i < A.length; ++i){
int result1 = maxArray[i - 1] * A[i];
int result2 = minArray[i - 1] * A[i];
maxArray[i] = Math.max(Math.max(result1, result2), A[i]);
minArray[i] = Math.min(Math.min(result1, result2), A[i]);
ans = Math.max(ans, maxArray[i]);
}
}
return ans;
}
}</span>
public int maxProduct(int[] A){
int ans = 0;
if(A != null && A.length != 0){
int maxArray[] = new int[A.length];
int minArray[] = new int[A.length];
ans = minArray[0] = maxArray[0] = A[0];
for(int i = 1; i < A.length; ++i){
int result1 = maxArray[i - 1] * A[i];
int result2 = minArray[i - 1] * A[i];
maxArray[i] = Math.max(Math.max(result1, result2), A[i]);
minArray[i] = Math.min(Math.min(result1, result2), A[i]);
ans = Math.max(ans, maxArray[i]);
}
}
return ans;
}
}</span>
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