Find the contiguous subarray within an array (containing at least one number) which has the largest sum.
For example, given the array [-2,1,-3,4,-1,2,1,-5,4]
,
the contiguous subarray [4,-1,2,1]
has the largest sum = 6
.
题目:给定一个数组,求连续最大的和。
思路一:一个个遍历,求所有的组合,复杂度是O(N^2),妥妥超时。
public int maxSubArray(int[] nums) {
int left = 0;
int right = nums.length - 1;
int max = result(nums,left,right);
for(int i = 0;i < nums.length;i++) {
int sum = 0;
for(int j = i;j <nums.length;j++){
sum+=nums[j];
if(sum > max) {
max = sum;
}
}
}
return max;
}
思路2:动态求和,如果前k个的和为sum,如果sum<0,那在加上k+1就没有意义,可以直接sum=0,重新开始统计。
public int maxSubArray(int[] nums) {
int sum = nums[0],max = nums[0];
for(int i = 1;i < nums.length;i++) {
if(sum < 0) sum = 0;
sum += nums[i];
if(sum > max) {
max = sum;
}
}
return max;
}