Container With Most Water (Java)
2015-01-03 21:21
239 查看
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.
a1,a2,......an是高度,它们之间的距离是底长,容器的面积是两个高度中最短的那个边乘以底长。找这个数列中能使容器面积最大的。设两个指针,一个在最左,一个在最右,都往中间走。判断条件是:如果height[low] < height[high] 那么 low++, 否则high--。 因为如果height[low]
< height[high],那么面积由low决定,这时high往左走,面积是不会增大的(不仅底长减小,而且要么high-1位置上的数大于low位置上的数面积由low决定,要么high-1位置上的数小于low位置上的数由此位决定,总之面积都是变小),所以唯一面积可能增大的方法就是low++。
Source
public int maxArea(int[] height) {
if(height.length == 0) return 0;
int max = 0;
int cap = 0;
int low = 0, high = height.length - 1;
while(low < high){
cap = (high - low) * Math.min(height[low], height[high]);
if(max < cap) max = cap;
if(height[low] < height[high]){
low ++;
}
else{
high --;
}
}
return max;
}
Test
public static void main(String[] args){
int[] height = {3,2,6,4,7,9};
int b = new Solution().maxArea(height);
System.out.println(b);
}
..., 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.
a1,a2,......an是高度,它们之间的距离是底长,容器的面积是两个高度中最短的那个边乘以底长。找这个数列中能使容器面积最大的。设两个指针,一个在最左,一个在最右,都往中间走。判断条件是:如果height[low] < height[high] 那么 low++, 否则high--。 因为如果height[low]
< height[high],那么面积由low决定,这时high往左走,面积是不会增大的(不仅底长减小,而且要么high-1位置上的数大于low位置上的数面积由low决定,要么high-1位置上的数小于low位置上的数由此位决定,总之面积都是变小),所以唯一面积可能增大的方法就是low++。
Source
public int maxArea(int[] height) {
if(height.length == 0) return 0;
int max = 0;
int cap = 0;
int low = 0, high = height.length - 1;
while(low < high){
cap = (high - low) * Math.min(height[low], height[high]);
if(max < cap) max = cap;
if(height[low] < height[high]){
low ++;
}
else{
high --;
}
}
return max;
}
Test
public static void main(String[] args){
int[] height = {3,2,6,4,7,9};
int b = new Solution().maxArea(height);
System.out.println(b);
}
相关文章推荐
- 【Leetcode】Container With Most Water in JAVA
- [LeetCode][Java] Container With Most Water
- LeetCode: Container With Most Water [Java]
- Container With Most Water (Java实现)
- leetcode Container With Most Water(Java)
- 【JAVA、C++】LeetCode 011 Container With Most Water
- [LeetCode][11]Container With Most Water解析 时间复杂度为O(n) -Java实现
- 【小熊刷题】Container with Most Water <Leetcode 11, Java>
- Container With Most Water leetcode java
- LeetCode 11 Container With Most Water (C,C++,Java,Python)
- leetcode:Container With Most Water 【Java】
- Leet Code 11 Container With Most Water - Java
- Container With Most Water leetcode java
- leetcode-java.T011_ContainerWithMostWater 找两条竖线然后这两条线以及X轴构成的容器能容纳最多的水
- (java)Container With Most Water
- 【LeetCode-面试算法经典-Java实现】【011-ContainerWithMostWater(容纳最多的水)】
- Java [leetcode 11] Container With Most Water
- java leetode Container With Most Water
- Java实现Container With Most Water
- [Leetcode] Container With Most Water (Java)