1. 题目
2. 思路
(1) 动态规划
- 遍历数组，计算每个以元素i为结尾的子区间的最大和，记录最大值返回。
- 每个以元素i为结尾的子区间的最大和，要么是以元素i-1为结尾的子区间的最大和加上元素i的值，要么就是元素i的值。
(2) 分治法
3. 代码
public class Test {
public static void main(String[] args) {
}
}
class Solution {
public int maxSubArray(int[] nums) {
int pre = nums[0];
int res = pre;
for (int i = 1; i < nums.length; i++) {
pre = Math.max(pre + nums[i], nums[i]);
res = Math.max(res, pre);
}
return res;
}
}
class Solution1 {
class Status {
public int lSum, rSum, mSum, iSum;
public Status(int lSum, int rSum, int mSum, int iSum) {
this.lSum = lSum;
this.rSum = rSum;
this.mSum = mSum;
this.iSum = iSum;
}
}
public int maxSubArray(int[] nums) {
return getInfo(nums, 0, nums.length - 1).mSum;
}
public Status getInfo(int[] a, int l, int r) {
if (l == r) {
return new Status(a[l], a[l], a[l], a[l]);
}
int m = (l + r) >> 1;
Status lSub = getInfo(a, l, m);
Status rSub = getInfo(a, m + 1, r);
return pushUp(lSub, rSub);
}
public Status pushUp(Status l, Status r) {
int iSum = l.iSum + r.iSum;
int lSum = Math.max(l.lSum, l.iSum + r.lSum);
int rSum = Math.max(r.rSum, r.iSum + l.rSum);
int mSum = Math.max(Math.max(l.mSum, r.mSum), l.rSum + r.lSum);
return new Status(lSum, rSum, mSum, iSum);
}
}