Continuous Subarray Sum II
Given an circular integer array (the next element of the last element is the first element), find a continuous subarray in it, where the sum of numbers is the biggest. Your code should return the index of the first number and the index of the last number.
If duplicate answers exist, return any of them.
Example
Give [3, 1, -100, -3, 4], return [4,1].
先是一个超时了的代码。。虽然剪了枝但还是会超时。
Time Limit Exceeded 91% test cases passed. Total Runtime: 12957 ms
public class Solution {
/**
* @param A an integer array
* @return A list of integers includes the index of the first number and the index of the last number
*/
public ArrayList<Integer> continuousSubarraySumII(int[] A) {
int l = A.length;
int s = 0;
int e = 0;
int ms = 0;
int me = 0;
int sum = A[0];
int max = A[0];
ArrayList<Integer> list = new ArrayList();
if(l == 1){
list.add(0);
list.add(0);
return list;
}
for(int j = 0;j<l;j++) {
if(A[j] <= 0) continue;
sum = A[j];
s = j;
e = j;
for(int i = (j+1)%l;j!=i;i = (i+1)%l){
if(A[i] < 0) {
if(sum > max) {
max = sum;
ms = s;
me = e;
}
}
if(sum < 0) {
sum = A[i];
s = i;
e = i;
}else {
sum += A[i];
e = i;
}
}
}
if(max > sum) {
list.add(ms);
list.add(me);
}else {
list.add(s);
list.add(e);
}
return list;
}
}
然后很仔细的想了一下。设0<=a<=b<=A.length,最大和区间只有两种情况:[a,b]和[b,A.length-1]U[0,a];只要找出这两种情况下的最大区间和，比较一下大的那个就是所求区间。
第一种情况就是求不头尾相连数组A的最大和区间。第二种是求不头尾相连A的最小子区间[a+1,b-1]（或者说是-A的最大区间和）。
敲出一段乱糟糟的代码终于AC了，容我去撸一把。
Accepted Total Runtime: 12746 ms 100% test cases passed.
public class Solution {
/**
* @param A an integer array
* @return A list of integers includes the index of the first number and the index of the last number
*/
public ArrayList<Integer> continuousSubarraySumII(int[] A) {
int[] nums = A;
int sum = nums[0];
int allsum = nums[0];
int max = nums[0];
int min = nums[0];
int msum = nums[0];
int mins = 0;
int mine = 0;
int s1 = 0;
int e1 = 0;
int s = 0;
int e = 0;
int ms = 0;
int me = 0;
ArrayList<Integer> list = new ArrayList<Integer>();
for(int i = 1 ;i < nums.length ;i++) {
allsum += nums[i];
if(nums[i] < 0) {
if(sum > max) {
max = sum;
ms = s;
me = e;
}
}
if(nums[i] > 0) {
if(msum < min) {
min = msum;
mins = s1;
mine = e1;
}
}
if(sum < 0) {
sum = nums[i];
s = i;
e = i;
}else {
sum+=nums[i];
e++;
}
if(msum > 0) {
msum = nums[i];
s1 = i;
e1 = i;
}else {
msum+=nums[i];
e1++;
}
}
if(sum > max) {
max = sum;
ms = s;
me = e;
}
if(msum < min) {
min = msum;
mins = s1;
mine = e1;
}
int val = allsum - min;
if(val > max && (mins != 0 && mine != A.length-1)) {
ms = mine + 1;
me = mins - 1;
}
list.add(ms);
list.add(me);
return list;
}
}
还有一个想法是将A拷贝一份接上，然后相当于求这整个数组的最大和区间。想法是设置一个int flag允许遍历数组尾部两次；同时限制区间长度不能超过A.length。但是WA了。快折腾一下午了实在不想搞了先放着以后想起来再说把。
Wrong Answer Total Runtime: 8168 ms 78% test cases passed.
public class Solution {
/**
* @param A an integer array
* @return A list of integers includes the index of the first number and the index of the last number
*/
public ArrayList<Integer> continuousSubarraySumII(int[] A) {
int l = A.length;
int s = 0;
int e = 0;
int ms = 0;
int me = 0;
int sum = A[0];
int max = A[0];
int flag = 2;//允许遍历数组尾部2次
ArrayList<Integer> list = new ArrayList();
if(l == 1){
list.add(0);
list.add(0);
return list;
}
int i = 1;
while(flag > 0) {
if(i == l - 1) flag--;
if(A[i] < 0) {
if(sum > max) {
max = sum;
ms = s;
me = e;
}
}
if(sum < 0) {
sum = A[i];
s = i;
e = 0;
}else {
if(e + 1 > l - 1) break;
sum += A[i];
e++;
}
i = (i + 1) % l;
}
if(max > sum) {
list.add(ms);
list.add((me+ms)%l);
}else {
list.add(s);
list.add((e+s)%l);
}
return list;
}
}