1.最小路径和
给定一个包含非负整数的
m x n
网格grid
,请找出一条从左上角到右下角的路径,使得路径上的数字总和为最小。说明:每次只能向下或者向右移动一步。
64. 最小路径和 - 力扣(LeetCode)
class Solution {public int minPathSum(int[][] grid) {if (grid == null || grid.length == 0 || grid[0].length == 0) {return 0;}int rows = grid.length, columns = grid[0].length;int[][] dp = new int[rows][columns];dp[0][0] = grid[0][0];for (int i = 1; i < rows; i++) {dp[i][0] = dp[i - 1][0] + grid[i][0];}for (int j = 1; j < columns; j++) {dp[0][j] = dp[0][j - 1] + grid[0][j];}for (int i = 1; i < rows; i++) {for (int j = 1; j < columns; j++) {dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j];}}return dp[rows - 1][columns - 1];}
}
2.加一
66. 加一 - 力扣(LeetCode)
class Solution {public int[] plusOne(int[] digits) {int n = digits.length;for (int i = n - 1; i >= 0; i--) {if (digits[i] != 9) {digits[i]++;for (int j = i + 1; j < n; j++) {digits[j] = 0;}return digits;}}// digits 中所有的元素均为 9int[] ans = new int[n + 1];ans[0] = 1;return ans;}
}
3.二进制求和
67. 二进制求和 - 力扣(LeetCode)
方法一:转化为十进制数
先将 aaa 和 bbb 转化成十进制数,求和后再转化为二进制数。利用 Python 和 Java 自带的高精度运算,我们可以很简单地写出这个程序:
class Solution {public String addBinary(String a, String b) {return Integer.toBinaryString(Integer.parseInt(a, 2) + Integer.parseInt(b, 2));}
}
方法二:模拟
class Solution {public String addBinary(String a, String b) {StringBuffer ans = new StringBuffer();int n = Math.max(a.length(), b.length()), carry = 0;for (int i = 0; i < n; ++i) {carry += i < a.length() ? (a.charAt(a.length() - 1 - i) - '0') : 0;carry += i < b.length() ? (b.charAt(b.length() - 1 - i) - '0') : 0;ans.append((char) (carry % 2 + '0'));carry /= 2;}if (carry > 0) {ans.append('1');}ans.reverse();return ans.toString();}
}
4.x的平方根
69. x 的平方根 - 力扣(LeetCode)
方法一:袖珍计算器算法
class Solution {public int mySqrt(int x) {if (x == 0) {return 0;}int ans = (int) Math.exp(0.5 * Math.log(x));return (long) (ans + 1) * (ans + 1) <= x ? ans + 1 : ans;}
}
方法二:二分查找
class Solution {public int mySqrt(int x) {int l = 0, r = x, ans = -1;while (l <= r) {int mid = l + (r - l) / 2;if ((long) mid * mid <= x) {ans = mid;l = mid + 1;} else {r = mid - 1;}}return ans;}
}
5.爬楼梯
70. 爬楼梯 - 力扣(LeetCode)
方法一:动态规划
class Solution {public int climbStairs(int n) {int p = 0, q = 0, r = 1;for (int i = 1; i <= n; ++i) {p = q; q = r; r = p + q;}return r;}
}