Container With Most Water (容器中盛最多的水)
2016-04-26 20:10
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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. 方法一:时间复杂度为o(n*n)。动态规划思想,每次选择当前最优解,然后用当前最优解和上一次的最优解比较。选择一个较大的解。此方法在leetcode中,时间越界。
int maxArea(vector<int>& height) { //最大容量 int max_val = 0; for (int i = 1; i < height.size(); ++i) { int val = 0; for (int j = 0; j < i; ++j) { //当前最大容量 val = (i - j)*min(height[i] , height[j]); if (val > max_val) max_val = val; } } return max_val; }
方法二:由于动态规划的时间复杂度为o(n*n),进一步我们可以找到时间复杂度为o(n)的算法,设置两个指针,初始化时分别指向首元素和尾元素。计算当前的最大容量。如果首元素小于尾元素,则首元素指针向前移动一位,直到大于尾元素。如果首元素大于尾元素,则尾元素指针向后移动一位,直到大于首元素为止,再次计算最大容量。
int maxArea(vector<int>& height) { //首元素指针 int i = 0; //尾元素指针 int j = height.size() - 1; //最大容量 int max_val = 0; while (i < j) { //当前最大容量 max_val = max(max_val, min(height[i], height[j])*(j - i)); //移动指针 height[i] < height[j] ? i++ : j--; } return max_val; }
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