[leetcode]11. Container With Most Water
2016-07-02 08:58
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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.
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分析:用两个指针从数组两端向中间查找,每次移动短板的那个指针,如果移动的是高的那个,此时装水量的高度仍然在短板,而且此时x轴长度还变短了,面积肯定减小;移动短的话,即使下一次高度仍然降低,通过比较ret中存放的依然是到目前为止面积最大的结果。
代码:
class Solution {
public:
int maxArea(vector<int>& height) {
int n=height.size();
if(n<2) return 0;
int ret=0;
int start=0;
int end=n-1;
while(start<end){
ret=max(ret,min(height[start],height[end])*(end-start));
if(height[start]<height[end]){
start++;
}
else{
end--;
}
}
return ret;
}
};
..., 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.
Subscribe to see which companies asked this question
分析:用两个指针从数组两端向中间查找,每次移动短板的那个指针,如果移动的是高的那个,此时装水量的高度仍然在短板,而且此时x轴长度还变短了,面积肯定减小;移动短的话,即使下一次高度仍然降低,通过比较ret中存放的依然是到目前为止面积最大的结果。
代码:
class Solution {
public:
int maxArea(vector<int>& height) {
int n=height.size();
if(n<2) return 0;
int ret=0;
int start=0;
int end=n-1;
while(start<end){
ret=max(ret,min(height[start],height[end])*(end-start));
if(height[start]<height[end]){
start++;
}
else{
end--;
}
}
return ret;
}
};
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