712.两个字符串的最小ASCII删除和

题目：
给定两个字符串s1 和 s2，返回 使两个字符串相等所需删除字符的 ASCII 值的最小和 。
算法：
方法一：动态规划
dp[i][j] 表示s1 取[0,i]个字符,s2 取[0,j]个字符，时两个字符串相等需要删除的字符ascii值最小和。
状态转移方程：
dp[i][j] = 0; i=0,j=0
dp[i][j] = dp[i][j - 1] + s2[j]; i=0,j!=0
dp[i][j] = dp[i - 1][j] + s1[i]; i=0,j!=0
dp[i][j] = dp[i - 1][j - 1] i!=0,j!=0,s1[i] == s2[j]
dp[i][j] = min(dp[i][j - 1] + s2[j], dp[i - 1][j] + s1[i]) i!=0,j!=0,s1[i] != s2[j]
现在就可以写代码了

func minimumDeleteSum(s1 string, s2 string) int {
	n1, n2 := len(s1), len(s2)
	dp := make([][]int, n1 + 1)
	for i := range dp {
		dp[i] = make([]int, n2 + 1)
	}
	for i := 0; i <= n1; i ++ {
		for j := 0; j <= n2; j ++ {
                        // dp[i][j]的index代表s1,s2取的字符最多个数
			s1Index := i - 1
			s2Index := j - 1
			if i == 0 && j != 0 {
				dp[i][j] = dp[i][j - 1] + int(s2[s2Index])
			} else if i != 0 && j == 0 {
				dp[i][j] = dp[i - 1][j] + int(s1[s1Index])
			} else if i != 0 && j != 0 {
				if s1[s1Index] == s2[s2Index] {
					dp[i][j] = dp[i - 1][j - 1]
				} else {
					dp[i][j] = min(dp[i][j - 1] + int(s2[s2Index]), dp[i - 1][j] + int(s1[s1Index]))
				}
			}
		}
	}
	return dp[n1][n2]
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}