本人的解法:
public class Main {
public int trap(int[] height) {
// 统计雨水总数//temp登记最大值
int sum = 0;
if (height.length != 0)
{
int temp = height[0];
int temp2 = 0;
int temp3 = height[0];
int temp4 = height[height.length - 1];
//跟踪最大值的位置
for (int x = 0; x < height.length; x++)
{
if (height[x] > temp) {
temp = height[x];
temp2 = x;
}
}
// 进行累加 这里采用两头向中间逐级判断
//第一个for
for (int i = 1; i <= temp2; i++)
{
if (height[i] > temp3)
{
temp3 = height[i];
}
else
{
sum = sum + temp3 - height[i];
}
}
//第二个for
for (int i = height.length - 1; i >= temp2; i--)
{
if (height[i] > temp4)
{
temp4 = height[i];
}
else
{
sum = sum + temp4 - height[i];
}
}
return sum;
}
//空集合
else
{
return 0;
}
}
//测试代码
public static void main(String[] args) {
Main m = new Main();
int[] arr = { 0, 1, 0, 2, 1, 0, 1, 3, 2, 1, 2, 1 };
System.out.println(m.trap(arr));
}
}
//////////////////////////////简化版
public class Solution {
public int trap(int[] A) {
int left = 0 , right = A.length-1;
int sum = 0;
int pre = 0;
while(left < right){
sum += (Math.min(A[left], A[right])-pre) * (right-left-1);
pre = Math.min(A[left],A[right]);
if(A[left] > A[right]){
int temp = right-1;
while(left < temp && A[temp] <= pre){sum-=A[temp];temp--;}
if(left < temp) sum -= pre;
right = temp;
}else{
int temp = left+1;
while(temp < right && A[temp] <= pre){sum-=A[temp];temp++;}
if(temp < right) sum -= pre;
left = temp;
}
}
return sum;
}
}
?
最大值的位置只选一个(如果几个最大值一样的话 也只选其中一个)
其他解法:
1、填充集合为凸集 填充的单位
2、
class Solution {
public int trap(int[] height) {
if(height.length<3){
return 0;
}
return find(height,0,height.length-1);
}
public int find(int[] height,int start,int end){
if(end-start<2){//递归的终点
return 0;
}
int max=-1,tmp=-1,min_two=Math.min(height[start],height[end]),sum=0;;
for(int i=start+1;i<end;i++){
//这一句写在哪里都行
sum=sum+(min_two-height[i]);
if(height[i]>max){
max=height[i];
tmp=i;
}
}
if(max<min_two){//上面的加法其实应该在这里,转移到上面和在这里其实都一样
//sum=sum+(min_two-height[i]);
return sum;
}else{//其实这里还可以优化一下当中间的max值等于start或者end的时候,当它等于start,那么直接计算即可,不用进行下一次递归,因为下一次递归会再扫描一遍
return find(height,start,tmp)+find(height,tmp,end);
}
}
}
public int trap(int[] A) {
if (A==null) return 0;
Stack<Integer> s = new Stack<Integer>();
int i = 0, maxWater = 0, maxBotWater = 0;
while (i < A.length){
if (s.isEmpty() || A[i]<=A[s.peek()]){
s.push(i++);
}
else {
int bot = s.pop();
maxBotWater = s.isEmpty()? // empty means no il
0:(Math.min(A[s.peek()],A[i])-A[bot])*(i-s.peek()-1);
maxWater += maxBotWater;
}
}
return maxWater;
}
原文地址:https://www.cnblogs.com/cznczai/p/11149678.html
时间: 2024-10-10 02:50:25