mysql 字符串函数(查询处理字符串)

Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.

For example,
Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.

对于每一个点来说，蓄水量 = min(左边的最高点，右边的最高点)-本身高度，于是两次遍历，一次用来找每个点的左边的最高点，一次用来找每个点的右边的最高点。

public class Solution {
public int trap(int[] A) {
int result = 0;
if(A.length<3){
return 0;
}
int []temp = new int[A.length];
int left = A[0];
for(int i=1;i<A.length-1;i++){
if(A[i]<left){
temp[i] = left;
}else if(A[i]>left){
left = A[i];
}
}
int right = A[A.length-1];
for(int i=A.length-2;i>0;i--){
if(A[i]<right){
if(temp[i]>0){
result += (min(temp[i],right)-A[i]);
}
}else if(A[i]>right){
right = A[i];
}
}
return result;
}

public int min(int a,int b){
return a>b?b:a;
}
}

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