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MaximumSubarray.java
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43 lines (37 loc) · 1.02 KB
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package LeetCodeOJ;
public class MaximumSubarray {
public int maxSubArray(int[] nums) {
//O(n^2)
// int Max = Integer.MIN_VALUE;
// for (int i=0;i< nums.length;i++){
// //记录每次从i开始的,j结束的 连续和
// int tmpMax = 0;
// for (int j=i;j<nums.length;j++){
// tmpMax += nums[j];
// //每做一次加法,与max比较,并取最大值
// if (tmpMax > Max){
// Max = tmpMax;
// }
// }
// }
// reutrn Max;
if (nums.length == 0){
return 0;
}
int Max = nums[0];
int conSum = Max;
for (int i=1; i< nums.length;i++){
//如果 conSum + nums[i] 比 nums[i] 还要小, 说明前面的和可以丢弃,直接使用nums[i]
conSum = conSum + nums[i] > nums[i]? conSum + nums[i]:nums[i];
if (conSum > Max){
Max = conSum;
}
}
return Max;
}
public static void main(String[] args) {
MaximumSubarray max = new MaximumSubarray();
int[]nums = { -2,1,-3,4,-1,2,1,-5,4};
System.out.println(max.maxSubArray(nums));
}
}