Title Description
HZ occasionally get some professional issues to flicker those non-computer science students. After the test group will finish today, he spoke up: in the old one-dimensional pattern recognition, it is often necessary to calculate the maximum and continuous sub-vector when the vector is a positive whole number, the problem solved. However, if the vector contains a negative number, it should contain a negative number, and expect a positive number next to it would make up for it? For example: {6, -3, -2,7, -15,1,2,2}, and the maximum successive sub-vectors of 8 (beginning from 0, up to the third). To an array and returns its maximum continuous subsequence and you will not be fooled him live? (Sub-vector length is at least 1)
Wrong Solutions:
public class FindGreatestSumOfSubArray {
public int FindGreatestSumOfSubArray(int[] array) {
int result=0;
if (array.length==1) {
return array[0];
}
else {
for (int i = 0; i < array.length; i++) {
for (int j = 0; j < array.length; j++) {
int sum=0;
while(i<j) {
sum+=array[i];
}
if (sum>result) {
result=sum;
}
}
}
}
return result;
}
Correct problem-solving ideas:
Using dynamic programming
F (i): in array [i] is the last element of the subarray and the maximum value, the relative position of the elements of the same subarray
F（i）=max（F（i-1）+array[i] ， array[i]）
res: All sub-arrays and the maximum
res=max（res，F（i））
The array of [6, -3, -2, 7, -15, 1, 2, 2]
Initial state:
F（0）=6
res=6
i=1：
F（1）=max（F（0）-3，-3）=max（6-3，3）=3
res=max（F（1），res）=max（3，6）=6
i=2：
F（2）=max（F（1）-2，-2）=max（3-2，-2）=1
res=max（F（2），res）=max（1，6）=6
i=3：
F（3）=max（F（2）+7，7）=max（1+7，7）=8
res=max（F（2），res）=max（8，6）=8
i=4：
F（4）=max（F（3）-15，-15）=max（8-15，-15）=-7
res=max（F（4），res）=max（-7，8）=8
And so on
The final value of 8 res
public int FindGreatestSumOfSubArray(int[] array) { int res = array[0]; //记录当前所有子数组的和的最大值 int max=array[0]; //包含array[i]的连续数组最大值 for (int i = 1; i < array.length; i++) { max=Math.max(max+array[i], array[i]); res=Math.max(max, res); } return res; }