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0053_maximum_subarray.java
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60 lines (52 loc) · 1.4 KB
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/*
给定一个整数数组 nums ，找到一个具有最大和的连续子数组（子数组最少包含一个元素），返回其最大和。
示例:
输入: [-2,1,-3,4,-1,2,1,-5,4],
输出: 6
解释: 连续子数组 [4,-1,2,1] 的和最大，为 6。
进阶:
如果你已经实现复杂度为 O(n) 的解法，尝试使用更为精妙的分治法求解。
*/
class MySolution {
public int maxSubArray(int[] nums) {
int max = Integer.MIN_VALUE;
int sum = 0;
for (int i = 0; i < nums.length; i++) {
for (int j = i; j < nums.length; j++) {
sum += nums[j];
if (sum > max)
max = sum;
}
sum = 0;
}
return max;
}
}
class Solution1 {
public int maxSubArray(int[] nums) {
int max = nums[0];
int sum = nums[0];
for (int i = 1; i < nums.length; i ++) {
sum = Math.max(nums[i], sum + nums[i]);
max = Math.max(max, sum);
}
return max;
}
}
class Solution2 {
public int maxSubArray(int[] nums) {
int sum = 0;
int max = Integer.MIN_VALUE;
for (int i = 0; i < nums.length; i++) {
if(sum + nums[i] < nums[i]){
sum = nums[i];
}else{
sum += nums[i];
}
if(sum > max){
max = sum;
}
}
return max;
}
}