Algorithms—42.Trapping Rain Water

public class Solution {
public int trap(int[] height) {
if (height.length<=1) {
return 0;
}
int a=0;
int h=height[a];
/**
* 求出最高峰
*/
for (int i = 1; i < height.length; i++) {
if (height[i]>h) {
a=i;
h=height[i];
}
}
Map<String, Integer> leftMap=new HashMap<String, Integer>();
Map<String, Integer> rightMap=new HashMap<String, Integer>();
leftMap.put("result", 0);
leftMap.put("a",a);
rightMap.put("result", 0);
rightMap.put("a",a);
Map<String, Integer> leftrRsult=new Solution().left(leftMap, height);
Map<String, Integer> rightResult=new Solution().right(leftMap, height);
return leftrRsult.get("result")+rightResult.get("result");
}

/**
* 求左侧
* @param height
* @param a
* @return
*/
public Map<String, Integer> left(Map<String, Integer> m,int[] height){
Map<String, Integer> map=new HashMap<String, Integer>();
if (m==null||m.get("a")==null) {
return m;
}
if (m.get("a")<=1) {
return m;
}
int h=height[0];
int result=m.get("result");
int sum=0;
int a=0;
for (int i = 1; i <m.get("a"); i++) {
if (height[i]<h) {
sum+=(h-height[i]);
}else {
sum=0;
h=height[i];
a=i;
}
}
result+=sum;
map.put("result", result);
map.put("a", a);

return left(map, height);
}
/**
* 求右侧
* @param height
* @param a
* @return
*/
public Map<String, Integer> right(Map<String, Integer> m,int[] height){
Map<String, Integer> map=new HashMap<String, Integer>();
if (m==null||m.get("a")==null) {
return m;
}
if (m.get("a")>=height.length-2) {
return m;
}
int h=height[height.length-1];
int result=m.get("result");
int sum=0;
int a=0;
for (int i = height.length-1; i>m.get("a"); i--) {
if (height[i]<h) {
sum+=(h-height[i]);
}else {
sum=0;
h=height[i];
a=i;
}
}
result+=sum;
map.put("result", result);
map.put("a", a);

return right(map, height);
}
}

思路：先找出最高峰，然后分为左右两个部分，以左边为例说明，从开始端作为参照开始往右遍历，如果遇到小于参照值的，那么加上参照值与当前值的差值，否则更换参照值；求出靠近最高峰最大值之间的蓄水量，然后递归进行再次求解。右边同理。