题解——Leetcode 11.Container With Most Water 难度:Medium
2017-04-02 20:55
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Given n non-negative integers a1,
a2, ..., an, where each represents a point at coordinate (i,
ai). n vertical lines are drawn such that the two endpoints of line
i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.
Note: You may not slant the container and n is at least 2.
C++程序如下:
class Solution {
public:
int maxArea(vector<int>& height) {
int begin = 0;
int end = height.size()-1;
int maxarea = 0;
while(begin<end){
maxarea = max(maxarea, min(height[begin], height[end])*(end-begin));
if(height[begin] < height[end])
begin++;
else
end--;
}
return maxarea;
}
};
任取两条线段,其与X轴合围的区域看作一个木桶,求木桶最大装水量。
木桶的装水量由最短的线段决定,所以装水量maxarea=min(height[begin], height[end])*(end-begin)。
程序从最左的线段和最右的线段同时开始遍历,左右线段之间的距离随之减小,由于装水量很大程度上取决于左右线段之间的距离,所以要设法找到比左右线段中较短的那条高的线段,如果左边线段较短,就左边开始遍历,直到找到一条线段比右边线段还长,此时就须从右边开始遍历,以此类推,直到左右线段重合。
在遍历过程中不断记录合围木桶的装水量,与之前记录的最大装水量比较,遍历完成后也就找到了最大的装水量。
a2, ..., an, where each represents a point at coordinate (i,
ai). n vertical lines are drawn such that the two endpoints of line
i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.
Note: You may not slant the container and n is at least 2.
C++程序如下:
class Solution {
public:
int maxArea(vector<int>& height) {
int begin = 0;
int end = height.size()-1;
int maxarea = 0;
while(begin<end){
maxarea = max(maxarea, min(height[begin], height[end])*(end-begin));
if(height[begin] < height[end])
begin++;
else
end--;
}
return maxarea;
}
};
任取两条线段,其与X轴合围的区域看作一个木桶,求木桶最大装水量。
木桶的装水量由最短的线段决定,所以装水量maxarea=min(height[begin], height[end])*(end-begin)。
程序从最左的线段和最右的线段同时开始遍历,左右线段之间的距离随之减小,由于装水量很大程度上取决于左右线段之间的距离,所以要设法找到比左右线段中较短的那条高的线段,如果左边线段较短,就左边开始遍历,直到找到一条线段比右边线段还长,此时就须从右边开始遍历,以此类推,直到左右线段重合。
在遍历过程中不断记录合围木桶的装水量,与之前记录的最大装水量比较,遍历完成后也就找到了最大的装水量。
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