70. Climbing Stairs

You are climbing a staircase. It takes n steps to reach the top.

Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?

题目解析：

• 完全背包问题

代码：

class Solution {
    public int climbStairs(int n) {
        int[] dp = new int[n+1];
        dp[0] = 1;
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= 2; j++) {
                if (i - j >= 0) {
                    dp[i] += dp[i - j];
                }
            }
        }
        return dp[n];
    }
}

322. Coin Change

You are given an integer array coins representing coins of different denominations and an integer amount representing a total amount of money.

Return the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return -1.

You may assume that you have an infinite number of each kind of coin.

题目解析：

• 如果求组合数就是外层for循环遍历物品，内层for遍历背包。
• 如果求排列数就是外层for遍历背包，内层for循环遍历物品。

代码：

class Solution {
    public int coinChange(int[] coins, int amount) {
        if (amount == 0) return 0;
        int[] dp = new int[amount+1];
        dp[0] = 0;
        for (int i = 1; i <= amount; i++) {
            dp[i] = amount + 1;
        }
        for (int coin: coins) {
            for (int i = 1; i <= amount; i++) {
                if (i - coin >= 0) {
                    dp[i] = Math.min(dp[i], dp[i - coin] + 1);
                }
            }
        }
        return dp[amount] == amount + 1 ? -1 : dp[amount];
    }
}

279. Perfect Squares

Given an integer n, return the least number of perfect square numbers that sum to n.

A perfect square is an integer that is the square of an integer; in other words, it is the product of some integer with itself. For example, 1, 4, 9, and 16 are perfect squares while 3 and 11 are not.

题目解析：

• 转化成完全背包问题

代码：

class Solution {
    public int numSquares(int n) {
        int k = (int)Math.sqrt(n);
        int[] dp = new int[n + 1];
        dp[0] = 0;
        for (int i = 1; i <= n; i++) {
            dp[i] = n;
        }
        for (int i = 1; i <= k; i++) {
            for (int j = 1; j <= n; j++) {
                if (j - i*i >= 0) {
                    dp[j] = Math.min(dp[j], dp[j - i*i] + 1);
                }
            }
        }
        return dp[n];
    }
}

总结

• 如果求组合数就是外层for循环遍历物品，内层for遍历背包。
• 如果求排列数就是外层for遍历背包，内层for循环遍历物品。